全部更新到 http://blog.csdn.net/caicaiatnbu 有部分题目采用STL中的vector，string，stack，queue等，我的博文C++ STL系列有对其详细介绍。 PAT 1001
#include<stdio.h> int main() { //freopen("D://input.txt", "r", stdin); int n; while(scanf("%d", &n) != EOF) { int count = 0;//统计步数 while(n != 1) { if(n % 2 == 0) n = n / 2; else n = (3*n + 1) / 2; count ++; } printf("%d\n", count); } return 0; }PAT 1002
#include <stdio.h> #include <string.h> int main() { //freopen("D://input.txt", "r", stdin); char s[101]; char numToChinese[10][6] = {"ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu"}; while(scanf("%s", &s) != EOF) { //printf("%s\n", s); //printf("%d\n", strlen(s)); int len = strlen(s); int sum = 0; for(int i = 0; i < len; i ++) { sum = sum + s[i] - '0'; } //printf("%d\n", sum); if(sum / 100 != 0) printf("%s %s %s\n", numToChinese[sum/100], numToChinese[sum/10], numToChinese[sum]); if(sum / 100 == 0 && sum / 10 % 10 != 0) printf("%s %s", numToChinese[sum/10], numToChinese[sum]); if(sum / 100 == 0 && sum / 10 % 10 == 0) printf("%s\n", numToChinese[sum]); } return 0; }PAT 1003
//首先满足条件：所有的字符只能由P、A、T三个构成。 //用count_p表示：P之前A的个数，用count_a表示：P和T之间A的个数，用count_t表示：T之后A的个数，需要满足count_p*count_a == count_t并且count_a >= 1 #include<stdio.h> #include<string.h> int main() { freopen("D://input.txt", "r", stdin); int n; while(scanf("%d", &n) != EOF) { while(n --) { char s[102]; scanf("%s", s); int pos_p = -1, pos_t = -1; int count_p = 0, count_a = 0, count_t = 0; bool flag_p = true, flag_t = true; for(int i = 0; i < strlen(s); i ++) { if(s[i] == 'P' && flag_p) { pos_p = i; flag_p = false; } if(s[i] == 'T' && flag_t) { pos_t = i; flag_t = false; } } if((pos_p == -1 || pos_t == -1) || pos_p > pos_t) { printf("NO\n"); continue; } else { for(int i = 0; i < pos_p; i ++) { if(s[i] == 'A') count_p ++; } for(int i = pos_p+1; i < pos_t; i ++) { if(s[i] == 'A') count_a ++; } for(int i = pos_t+1; i < strlen(s); i ++) { if(s[i] == 'A') count_t ++; } if((count_p + count_a + count_t + 2) != strlen(s) || count_a == 0) { printf("NO\n"); continue; } } if((count_p * count_a) == count_t) printf("YES\n"); else printf("NO\n"); } } return 0; }PAT 1004
//只需要遍历一遍，不需要进行排序。 /* #include<stdio.h> #include<string.h> int main() { //freopen("D://input.txt", "r", stdin); int n; while(scanf("%d", &n) != EOF) { char name_min[11], id_min[11];//保存分数最低学生的姓名和学号 char name_max[11], id_max[11];//保存分数最高学生的姓名和学号 int score_min = 100, score_max = 0;//保存分数最高学生的分数 和 分数最高学生的分数 while(n --) { char name[11], id[11]; int score; scanf("%s %s %d", &name, &id, &score); printf("%s %s %d\n", name, id, score); if(score > score_max) { strcpy(name_max, name); strcpy(id_max, id); score_max = score; } if(score < score_min) { strcpy(name_min, name); strcpy(id_min, id); score_min = score; } } printf("%s %s\n", name_max, id_max); printf("%s %s\n", name_min, id_min); } return 0; }PAT 1005
// 用count[i] = 1表示此输入的i是关键数据，count[i] = 2表示此i不是关键数据，已经在被覆盖了 #include<stdio.h> #include<string.h> int count[102]; int main() { //freopen("D://input.txt", "r", stdin); int n; while(scanf("%d", &n) != EOF) { memset(count, 0, sizeof(count)); for(int i = 0; i < n; i ++) { int tmp; scanf("%d", &tmp); count[tmp] = 1; } for(int i = 1; i < 101; i ++) { if(count[i] != 1) continue; int tmp = i; while(tmp != 1) { if(tmp % 2 == 0) tmp = tmp / 2; else tmp = (3*tmp + 1) / 2; if(tmp <= 100) count[tmp] = 2; } } bool flag = true; for(int i = 100; i > 1; i --) { if(count[i] == 1) { if(flag) { printf("%d", i); flag = false; } else printf(" %d", i); } } printf("\n"); } return 0; }PAT 1006
#include<stdio.h> int main() { //freopen("D://input.txt", "r", stdin); int n; while(scanf("%d", &n) != EOF) { int a = n / 100; int b = n / 10 % 10; int c = n % 10; if(a > 0) { for(int i = 0; i < a; i ++) printf("B"); } if(b > 0) { for(int i = 0; i < b; i ++) printf("S"); } if(c > 0) { for(int i = 1; i <= c; i ++) printf("%d", i); } printf("\n"); } return 0; }PAT 1007
// 从第一个素数2开始，判断i和i+2是否同时为素数，如果同时为素数，则累加器count++，一直循环到i+2 <= n, 步长可选2 #include<stdio.h> #include<math.h> bool isPrime(int x) { for(int i = 2; i*i <= x; i ++) { if(x % i == 0) return false; } return true; } int main() { //freopen("D://input.txt", "r", stdin); int n; while(scanf("%d", &n) != EOF) { int count = 0; for(int i = 2; i <= n-2; i ++) { if(isPrime(i) && isPrime(i+2)) count ++; } printf("%d\n", count); } return 0; }PAT 1008
// 数组向右移动的值m可能大于数组的长度n，故需要对m = m % n; // 假设数组里边有3个元素，依次为1、2、3，我们将数组变为：1、2、3、1、2、3；这样向右移动遍历一遍就出来，从第n-m个打印到2*n-m #include<stdio.h> int main() { //freopen("D://input.txt", "r", stdin); int buf[202]; int n, m; while(scanf("%d", &n) != EOF) { scanf("%d", &m); for(int i = 0; i < n; i ++) { scanf("%d", &buf[i]); buf[n+i] = buf[i]; } m = m % n; for(int i = n-m; i < 2*n- m; i ++) { if(i == n-m) printf("%d", buf[i]); else printf(" %d", buf[i]); } printf("\n"); } return 0; }PAT 1009
#include<stdio.h> #include<iostream> #include<vector> using namespace std; int main() { //freopen("D://input.txt", "r", stdin); string s; vector<string> v; while(cin>>s) { v.push_back(s); } for(int i = v.size()-1; i >= 0; i --) if(i != 0) cout<<v[i]<<" "; else cout<<v[i]<<endl; return 0; }PAT 1010
// 如果某项中系数或者指数有一个为0，该项求导后为0，不需要显示； // 同时，如果像 0 2 0 1 1 0 这样的输入，求完导整体为0，此时我们需要显示0 0， // 这个case容易被忽视 #include<stdio.h> #include<iostream> #include<vector> using namespace std; int main() { freopen("D://input.txt", "r", stdin); vector<int> v, res; v.clear(); res.clear(); bool first = true; int n, m; while(scanf("%d", &n) != EOF) { v.push_back(n); } for(int i = 0; i < v.size(); i = i+2) { if(v[i] != 0 && v[i+1] != 0) { res.push_back(v[i]*v[i+1]); res.push_back(v[i+1]-1); } } if(res.size() == 0) printf("0 0\n"); else { for(int i = 0; i < res.size(); i += 2) { if(i == 0) printf("%d %d", res[i], res[i+1]); else printf(" %d %d", res[i], res[i+1]); } printf("\n"); } return 0; }PAT 1011
// 要用long long数据类型，用int类型会导致数据溢出。 #include<stdio.h> int main() { int n; //freopen("D://input.txt", "r", stdin); while(scanf("%d", &n) != EOF) { int i = 1; while(n --) { long long a, b, c; scanf("%lld %lld %lld", &a, &b, &c); if(a + b > c) printf("Case #%d: true\n", i); else printf("Case #%d: false\n", i); i ++; } } return 0; }PAT 1012
#include<stdio.h> int main() { //freopen("D://input.txt", "r", stdin); int n, buf[1002]; while(scanf("%d", &n) != EOF) { for(int i = 0; i < n; i ++) scanf("%d", &buf[i]); int A1 = 0, A2 = 0, A3 = 0, A5 = 0; double A4 = 0.0; int count1 = 0, count2 = 0, count3 = 0, count4 = 0, count5 = 0; bool flag = true; for(int i = 0; i < n; i ++) { if(buf[i] % 5 == 0 && buf[i] % 2 == 0) { A1 += buf[i]; count1 = 1; } if(buf[i] % 5 == 1) { if(flag) A2 += buf[i]; else A2 -= buf[i]; flag = !flag; count2 = 1; } if(buf[i] % 5 == 2) { A3 ++; count3 = 1; } if(buf[i] % 5 == 3) { A4 += buf[i]; count4 ++; } if(buf[i] % 5 == 4) { if(buf[i] > A5) A5 = buf[i]; count5 = 1; } } if(count1 == 0) printf("N "); else printf("%d ", A1); if(count2 == 0) printf("N "); else printf("%d ", A2); if(count3 == 0) printf("N "); else printf("%d ", A3); if(count4 == 0) printf("N "); else printf("%.1f ", A4*1.0/count4); if(count5 == 0) printf("N\n"); else printf("%d\n", A5); } return 0; }PAT 1013
#include<stdio.h> bool isPrime(int x) { for(int i = 2; i*i <= x; i ++) if(x % i == 0) return false; return true; } int main() { //freopen("D://input.txt", "r", stdin); int m, n; int buf[10002]; while(scanf("%d %d", &m, &n) != EOF) { int count = 1; int i = 2; while(count <= 10000) { if(isPrime(i)) { buf[count] = i; count ++; } i ++; } int len = 0; for(i = m; i <= n; i ++) { if(len == 0) printf("%d", buf[i]); else printf(" %d", buf[i]); len ++; if(len == 10) { len = 0; printf("\n"); } } //printf("\n"); } return 0; }PAT 1014
// 第一个字符串和第二个字符串，第一对相同的大写字母表示星期（A表示星期一，G表示星期天） // 第二对相对的数字和大写字母表示小时（0-9依次表示0-9点，A-N依次表示10-23点） // 在输出小时的时候，如果是8这样的，要输出成 08 // 第三个字符串和第四个串第一队相同的字母（A-Z，a—z）出现的位置就是分钟，就字符串角码。 // 也要输出成 02 这种格式 #include<stdio.h> #include<iostream> #include<string.h> using namespace std; int main() { //freopen("D://input.txt", "r", stdin); string Date[7] = {"MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"}; string s1, s2, s3, s4; bool flag = 1; while(cin>>s1>>s2>>s3>>s4) { char date ; int hour ; int minute ; int flag = 1; for(int i = 0; i < s1.length() && i < s2.length(); i ++) { if(s1[i] == s2[i] && s1[i] >= 'A' && s1[i] <= 'G' && flag == 1) { date = s1[i]; flag = 2; continue; //break; } if(s1[i] == s2[i] && ((s1[i] >= 'A' && s1[i] <= 'N') || (s1[i] >= '0' && s1[i] <= '9')) && flag == 2) { if(s1[i] >= 'A' && s1[i] <= 'N') { hour = s1[i] - 'A' + 10; break; } if(s1[i] >= '0' && s1[i] <= '9') { hour = s1[i] - '0'; break; } } } for(int i = 0; i < s3.length() && i < s4.length(); i ++) { if(s3[i] == s4[i] && ((s3[i] >= 'A' && s4[i] <= 'Z') || (s3[i] >= 'a' && s4[i] <= 'z'))) { minute = i; } } cout<<Date[date-'A']<<" "; printf("d:", hour); printf("d\n", minute); } return 0; }PAT 1015
// 结构数组排序，总共有四类，开辟4个vector来依次保存这四类。依次压入到v1，v2，v3，v4 // 第二个考点就在cmp函数的书写， #include<stdio.h> #include<iostream> #include<algorithm> #include<string.h> #include<vector> using namespace std; struct student { char ID[10]; int x;//德分 int y;//才分 }; bool cmp(const student &a, const student &b) { if((a.x+a.y) != (b.x+b.y)) return (a.x+a.y) > (b.x+b.y); else if(a.x != b.x) return a.x > b.x; else { for(int i = 0; i < strlen(a.ID) && i < strlen(b.ID); i ++) { if(a.ID[i] != b.ID[i]) return a.ID[i] < b.ID[i]; } } } int main() { //freopen("D://input.txt", "r", stdin); int n, L, H; vector<student> v1, v2, v3, v4; student tmp; while(scanf("%d", &n) != EOF) { v1.clear(); v2.clear(); v3.clear(); v4.clear(); scanf("%d %d", &L, &H); for(int i = 0; i < n; i ++) { scanf("%s %d %d", tmp.ID, &tmp.x, &tmp.y); if(tmp.x >= H && tmp.y >= H) v1.push_back(tmp); else if(tmp.x >= H && tmp.y >= L) v2.push_back(tmp); else if(tmp.x >= L && tmp.y >= L && tmp.x >= tmp.y) v3.push_back(tmp); else if(tmp.x >= L && tmp.y >= L) v4.push_back(tmp); } sort(v1.begin(), v1.end(), cmp); sort(v2.begin(), v2.end(), cmp); sort(v3.begin(), v3.end(), cmp); sort(v4.begin(), v4.end(), cmp); printf("%d\n", v1.size() + v2.size() + v3.size() + v4.size()); for(int i = 0; i < v1.size(); i ++) printf("%s %d %d\n", v1[i].ID, v1[i].x, v1[i].y); for(int i = 0; i < v2.size(); i ++) printf("%s %d %d\n", v2[i].ID, v2[i].x, v2[i].y); for(int i = 0; i < v3.size(); i ++) printf("%s %d %d\n", v3[i].ID, v3[i].x, v3[i].y); for(int i = 0; i < v4.size(); i ++) printf("%s %d %d\n", v4[i].ID, v4[i].x, v4[i].y); } return 0; }PAT 1016
#include<stdio.h> #include<string.h> int main() { //freopen("D://input.txt", "r", stdin); char a[12], b[12]; int Da, Db; while(scanf("%s %d %s %d", a, &Da, b, &Db) != EOF) { int sum1 = 0, sum2 = 0; for(int i = 0; i < strlen(a); i ++) { if((a[i] - '0') == Da) sum1 = sum1*10 + Da; } for(int i = 0; i < strlen(b); i ++) { if((b[i] - '0') == Db) sum2 = sum2*10 + Db; } printf("%d\n", sum1+sum2); } return 0; }PAT 1017
// 因为除数b是一位数，被除数是a[]，被除数从高位到底位逐位初以b， // 举例123除以5时候，首先考虑第一位，此时结果1/5 = 0余数1，再考虑第二位，此时先加上第一位的余数，即变为12/5=2余数2，考虑第三位时候，此时应为23/5=4余数为3，所以结果为024.....4,所以从第一非0打印结果，用c保存余数。 #include<stdio.h> #include<string.h> int main() { //freopen("D://input.txt", "r" stdin); char a[1002]; int res[10002]; int b; while(scanf("%s %d", a, &b) != EOF) { //memset(res, 0, sizeof(res)); int c = 0; if(strlen(a) == 1) { if((a[0]-'0') / b == 0) printf("0 %d\n", (a[0]-'0') % b); else printf("%d %d\n", (a[0]-'0') / b, (a[0]-'0') % b); continue; } for(int i = 0; i < strlen(a); i ++) { if((10*c+(a[i]-'0')) / b != 0) { res[i] = (10*c+(a[i]-'0')) / b; if((10*c+(a[i]-'0')) % b != 0) { c = (10*c+(a[i]-'0')) % b; } else c = 0; } else { res[i] = 0; if((10*c+(a[i]-'0')) % b != 0) { c = (10*c+(a[i]-'0')) % b; } else c = 0; } } bool first = true; for(int i = 0; i < strlen(a); i ++) { if(res[i] == 0 && first) { continue; } else { first = false; printf("%d", res[i]); } } printf(" %d\n", c); } return 0; }PAT 1018
// 甲赢得次数其实与乙输的次数相等，甲乙平手的次数相等。 #include<stdio.h> #include<string.h> int main() { //freopen("D://input.txt", "r", stdin); char a[2], b[2]; int n; while(scanf("%d", &n) != EOF) { int count_a[6];//count_a1中角码0表示甲胜的次数，1->甲平的次数，2->甲输的次数，3->甲赢出锤子，4->甲赢出布，5->甲赢出剪刀 int count_b[6]; memset(count_a, 0, sizeof(count_a)); memset(count_b, 0, sizeof(count_b)); for(int i = 0; i < n; i ++) { scanf("%s %s", a, b); if((a[0] == 'C' && b[0] == 'C') || (a[0] == 'J' && b[0] == 'J') || (a[0] == 'B' && b[0] == 'B')) { count_a[1] ++; count_b[1] ++; } else if((a[0] == 'B' && b[0] == 'C') || (a[0] == 'C' && b[0] == 'J') || (a[0] == 'J' && b[0] == 'B')) { count_a[0] ++; count_b[2] ++; if(a[0] == 'C') count_a[3] ++; else if(a[0] == 'B') count_a[4] ++; else count_a[5] ++; } else { count_a[2] ++; count_b[0] ++; if(b[0] == 'C') count_b[3] ++; else if(b[0] == 'B') count_b[4] ++; else count_b[5] ++; } } printf("%d %d %d\n", count_a[0], count_a[1], count_a[2]); printf("%d %d %d\n", count_b[0], count_b[1], count_b[2]); if(count_a[4] >= count_a[3] && count_a[4] >= count_a[5]) printf("B "); else if(count_a[3] >= count_a[4] && count_a[3] >= count_a[5]) printf("C "); else if(count_a[5] >= count_a[3] && count_a[5] >= count_a[4]) printf("J "); if(count_b[4] >= count_b[3] && count_b[4] >= count_b[5]) printf("B\n"); else if(count_b[3] >= count_b[4] && count_b[3] >= count_b[5]) printf("C\n"); else if(count_b[5] >= count_b[3] && count_b[5] >= count_b[4]) printf("J\n"); } return 0; }PAT 1019
// 输入的范围是（0，10000），根据输入要生成最大数max和最小数min。输入不为4位数，我们再生成max和min要补0，补成4位数。 // 其最终结果肯定是6174。 #include<stdio.h> #include<vector> #include<algorithm> using namespace std; vector<int> v; int max(int x) { int sum = 0; v.clear(); while(x != 0) { v.push_back(x); x = x / 10; } while(v.size() != 4) v.push_back(0); sort(v.begin(), v.end()); for(int i = v.size()-1; i >= 0; i --) sum = sum*10 + v[i]; return sum; } int min(int x) { int sum = 0; v.clear(); while(x != 0) { v.push_back(x); x = x / 10; } while(v.size() != 4) v.push_back(0); sort(v.begin(), v.end()); for(int i = 0; i < v.size(); i ++) sum = sum*10 + v[i]; return sum; } int main() { //freopen("D://input.txt", "r", stdin); int n; //while(scanf("%d", &n) != EOF) scanf("%d", &n); { if(n == 6174) { printf("d - d = d\n", max(n), min(n), max(n)-min(n)); n = max(n) - min(n); } while(n != 6174) { if(max(n) == min(n)) { printf("d - d = 0000\n", max(n), min(n)); break; } printf("d - d = d\n", max(n), min(n), max(n)-min(n)); n = max(n) - min(n); } } return 0; }PAT 1020
// 结构数组排序，按照月饼价格从高到低排序，价格最高的先卖，直到卖到m #include<stdio.h> #include<algorithm> using namespace std; struct moon { double x;//总量 double y;//总价 double price; }; bool cmp(const moon &a, const moon &b) { if(a.price != b.price) return a.price > b.price; else return a.price > b.price; } int main() { // freopen("D://input.txt", "r", stdin); int n, m; moon buf[1002]; while(scanf("%d %d", &n, &m) != EOF) { for(int i = 0; i < n; i ++) scanf("%lf", &buf[i].x); for(int i = 0; i < n; i ++) scanf("%lf", &buf[i].y); for(int i = 0; i < n; i ++) buf[i].price = buf[i].y/buf[i].x; sort(buf, buf+n, cmp); double sum = 0.0; for(int i = 0; i < n; i ++) { if(m > buf[i].x) { sum += buf[i].y; m = m - buf[i].x; } else { sum = sum + buf[i].price*m; break; } } printf("%.2f\n", sum); } return 0; }PAT 1021
#include<stdio.h> #include<string.h> int main() { //freopen("D://input.txt", "r", stdin); char s[1002]; int count[11]; while(scanf("%s", s) != EOF) { memset(count, 0, sizeof(count)); for(int i = 0; i < strlen(s); i ++) { count[s[i]-'0'] ++; } for(int i = 0; i < 10; i ++) { if(count[i] != 0) printf("%d:%d\n", i, count[i]); } } return 0; }PAT 1022
// int类型就可以，a和b的和不会溢出。考点就是把一个十进制数转为另一进制，取余法。 #include<stdio.h> #include<vector> using namespace std; int main() { //freopen("D://input.txt", "r", stdin); int a, b, d; while(scanf("%d %d %d", &a, &b, &d) != EOF) { int sum = a + b; vector<int> v; v.clear(); if(sum == 0) printf("0"); while(sum != 0) { v.push_back(sum%d); sum = sum / d; } for(int i = v.size()-1; i >=0; i --) { printf("%d",v[i]); } printf("\n"); } return 0; }PAT 1023
#include<stdio.h> #include<string.h> int main() { int buf[10]; memset(buf, 0, sizeof(buf)); for(int i = 0; i < 10; i ++) scanf("%d", &buf[i]); for(int i = 1; i < 10; i ++) { if(buf[i] != 0) { printf("%d", i); buf[i] --; break; } } for(int i = 0; i < 10; i ++) { for(int j = 0; j < buf[i]; j ++) printf("%d", i); } printf("\n"); return 0; }PAT 1024
#include<stdio.h> #include<string.h> int main() { //freopen("D://input.txt", "r", stdin); char s[10010]; while(scanf("%s", s) != EOF) { bool flag1, flag2; if(s[0] == '+') flag1 = true; else flag1 = false; char *buf1, *buf2, *buf3; const char *d = ".E"; buf1 = strtok(s, d); buf2 = strtok(NULL, d); buf3 = strtok(NULL, d); if(buf3[0] == '+') flag2 = true; else flag2 = false; int count = 0; for(int i = 1; i < strlen(buf3); i ++) count = count*10 + (buf3[i] - '0'); // case + if(!flag1) printf("-"); if(count == 0) { for(int i = 1; i < strlen(buf1); i ++) printf("%c", buf1[i]); printf(".%s", buf2); continue; } if(flag2) { for(int i = 1; i < strlen(buf1); i ++) printf("%c", buf1[i]); if(count < strlen(buf2)) { for(int i = 0; i < strlen(buf2); i ++) { if(i < count) printf("%c", buf2[i]); else if(i == count) { printf(".%c", buf2[i]); } else printf("%c", buf2[i]); } } else { for(int i = 0; i < count; i ++) { if(i < strlen(buf2)) printf("%c", buf2[i]); else printf("0"); } } } else { if(count < (strlen(buf1)-1)) { for(int i = 1; i < strlen(buf1); i ++) { if((count + i) == strlen(buf1)) printf(".%c", buf1[i]); else printf("%c", buf1[i]); } printf("%s", buf2); } else { printf("0."); for(int i = 0; i < (count+1-strlen(buf1)); i ++) printf("0"); for(int i = 1; i < strlen(buf1); i ++) printf("%c", buf1[i]); printf("%s", buf2); } } printf("\n"); } return 0; }PAT 1025
// 注意输入的节点有可能不在链表上； // 开辟一个buf[100002] 的结构数组，将data和next 根据其address保存到buf[address]中，然后根据firstaddress将链表上的节点加载到vector中，最后计算需要翻转的次数num = v.size()/k,用reverse(v.begin()+i*k, v.begin()+(i+1)*k)进行翻转,k从0到num #include<stdio.h> #include<iostream> #include<vector> #include<algorithm> using namespace std; struct node { int address; int data; int next; }buf[100002]; vector<node> v; int main() { freopen("D://input.txt", "r", stdin); int firstAddress, n, k; while(scanf("%d %d %d", &firstAddress, &n, &k) != EOF) { for(int i = 0; i < n; i ++) { int index; scanf("%d", &index); buf[index].address = index; scanf("%d %d", &buf[index].data, &buf[index].next); } if(firstAddress == -1) printf("-1\n"); else { int p = firstAddress; while(p != -1) { v.push_back(buf[p]); p = buf[p].next; } int countNode = v.size(); int num = countNode / k;//反转次数 因为k<=n 所以num>=1 for(int i = 0; i < num; i ++) std::reverse(v.begin() + i*k, v.begin() + (i+1)*k); for(int i = 0; i < countNode; i ++) { if((i+1) != countNode) printf("d %d d\n", v[i].address, v[i].data, v[i+1].address); else printf("d %d -1\n", v[i].address, v[i].data); } } } return 0; }PAT 1026
#include<stdio.h> int main() { freopen("D://input.txt", "r", stdin); int c1, c2; while(scanf("%d %d", &c1, &c2) != EOF) { int ans = (c2-c1) / 100; if((c2 - c1) % 100 >= 50) ans += 1; int h = ans / 3600; int m = (ans - h*3600) / 60; int s = ans % 60; printf("d:d:d\n", h, m, s); } return 0; }PAT 1027
#include<stdio.h> void print(int n, char s[]) { for(int i = 0; i < n; i ++) { for(int j = 0; j < n; j ++) { if((j >= i && (i+j) <= (n-1)) || (i >= j) && (i+j) >= (n-1)) printf("%s", s); else if(j < i && i <= n/2) printf(" "); else if((i+j) < (n-1) && i > n/2) printf(" "); } printf("\n"); } } int main() { freopen("D://input.txt", "r", stdin); int n; char s[2]; while(scanf("%d %s", &n, s) != EOF) { int level = -1; int count = 0; for(int i = 1; n >=0; i += 2) { if(i == 1) n = n - 1; else n = n - 2*i; if(n >= 0) { level += 2; count = n; } } print(level, s); printf("%d\n", count); //printf("level = %d count = %d\n", level, count); } return 0; }PAT 1028
// 注意满足条件的可能有0个人 #include<stdio.h> #include<string.h> int main() { freopen("D://input.txt", "r", stdin); int n; char name[6],birthday[12]; while(scanf("%d", &n) != EOF) { char max[12] = "2014/09/06"; char min[12] = "1814/09/06"; char name1[6], name2[6]; char limit1[12] = "1814/09/06"; char limit2[12] = "2014/09/06"; int count = 0; while(n --) { scanf("%s %s", name, birthday); if(strcmp(birthday, limit1) < 0) continue; if(strcmp(birthday, limit2) > 0) continue; if(strcmp(birthday, max) < 0) { strcpy(name1, name); strcpy(max, birthday); } if(strcmp(birthday, min) > 0) { strcpy(name2, name); strcpy(min, birthday); } count ++; } if(count != 0) printf("%d %s %s\n", count, name1, name2); else printf("0\n"); } return 0; }PAT 1029
#include<stdio.h> #include<string.h> int main() { //freopen("D://input.txt", "r", stdin); char s[82], ans[82]; int count[128]; while(scanf("%s %s", s, ans) != EOF) { memset(count, 0, sizeof(count)); char ch; for(int i = 0; i < strlen(ans); i ++) { count[ans[i]] = 1; if(ans[i] >= 'a' && ans[i] <= 'z') count[ans[i]-'a'+'A'] = 1; if(ans[i] >= 'A' && ans[i] <= 'Z') count[ans[i]-'A'+'a'] = 1; } for(int i = 0; i< strlen(s); i ++) { if(count[s[i]] == 0) { count[s[i]] = 2; if(s[i] >= 'a' && s[i] <= 'z') { printf("%c", s[i]-'a'+'A'); count[s[i]-'a'+'A'] = 2; continue; } printf("%c", s[i]); if(s[i] >= 'A' && s[i] <= 'Z') count[s[i]-'A'+'a'] = 2; } } printf("\n"); } return 0; }PAT 1030
// 注意：P是float #include<stdio.h> #include<vector> #include<algorithm> using namespace std; int main() { //freopen("D://input.txt", "r", stdin); int n; float p; vector<int> v; while(scanf("%d %f", &n, &p) != EOF) { int length = 1; v.clear(); int tmp; for(int i = 0; i < n; i ++) { scanf("%d", &tmp); v.push_back(tmp); } sort(v.begin(), v.end()); for(int i = 0; i < v.size(); i ++) { for(int j = i+length-1; j < v.size(); j ++) { if(v[j] <= v[i]*p) { if(j-i+1 > length) length = j - i + 1; } else break; } } printf("%d\n", length); } return 0; }PAT 1031
#include<stdio.h> #include<string.h> int main() { freopen("E://input.txt", "r", stdin); int n; char s[20]; char z[11] = {'1','0','X','9','8','7','6','5','4','3','2'}; int w[18] = {7, 9, 10, 5, 8, 4, 2, 1, 6, 3, 7, 9, 10, 5, 8, 4, 2}; while(scanf("%d", &n) != EOF) { int count = 0; while(n --) { scanf("%s", s); int sum = 0; for(int i = 0; i < 17; i ++) sum = sum + (s[i] - '0')*w[i]; sum = sum % 11; if(s[17] != z[sum]) { printf("%s\n", s); count ++; } } if(count == 0) printf("All passed\n"); } return 0; }PAT 1032
#include<stdio.h> #include<string.h> int main() { //freopen("D://input.txt", "r", stdin); int n; int buf[100002]; while(scanf("%d", &n) != EOF) { int index, score; memset(buf, 0, sizeof(buf)); for(int i = 0; i < n; i ++) { scanf("%d %d", &index, &score); buf[index] += score; } index = -1; score = 0; for(int i = 0; i <= 100000; i ++) { if(buf[i] > score) { index = i; score = buf[i]; } } printf("%d %d\n", index, score); } return 0; }PAT 1033
// 注意：第一行输入可能是为空，是一个空行，故需要用gets()来获取第一行的输入 #include<stdio.h> #include<string.h> int main() { //freopen("D://input.txt", "r", stdin); char error[100002]; bool flag[128]; char ch; gets(error); memset(flag, 0, sizeof(flag)); for(int i = 0; i < strlen(error); i ++) { ch = error[i]; flag[ch] = 1; if(ch >= 'A' && ch <= 'Z') { ch = ch - 'A' + 'a'; flag[ch] = 1; } } while(scanf("%c", &ch) != EOF) { if(flag[ch] == 0) { if(ch >= 'A' && ch <= 'Z' && flag['+'] == 1) continue; printf("%c", ch); } } return 0; }PAT 1034
// 注意在计算过程中，计算gcd和lcm可能会超出int表示的数据范围，所以得用long类型 #include<stdio.h> #include<string.h> long gcd(long x, long y) { return x == 0 ? y : gcd(y%x, x); } long lcm(long x, long y) { return x/gcd(x,y)*y; } void print(long x, long y) { bool flag = false; if(x < 0 ) { x = -x; flag = true; } if(x / y != 0 && x % y != 0) { if(flag) printf("(-%d %d/%d)", x / y, x%y/gcd(x%y,y), y/gcd(x%y,y)); else printf("%d %d/%d", x / y, x%y/gcd(x%y,y), y/gcd(x%y,y)); } else if(x / y == 0 && x % y != 0) { if(flag) printf("(-%d/%d)", x/gcd(x,y), y/gcd(x,y)); else printf("%d/%d", x/gcd(x,y), y/gcd(x,y)); } else if(x / y != 0 && x % y == 0) { if(flag) printf("(-%d)", x/y); else printf("%d", x/y); } else printf("0"); } int main() { freopen("D://input.txt", "r", stdin); long a, b, c, d; while(scanf("%ld/%ld %ld/%ld", &a, &b, &c, &d) != EOF) { // case + long res1 = 0, res2 = 0; print(a, b); printf(" + "); print(c, d); printf(" = "); res1 = a * lcm(b, d)/b + c * lcm(b, d)/d; res2 = lcm(b, d); print(res1, res2); printf("\n"); // case - print(a, b); printf(" - "); print(c, d); printf(" = "); res1 = a* lcm(b, d)/b - c* lcm(b, d)/d; res2 = lcm(b, d); print(res1, res2); printf("\n"); //case * print(a, b); printf(" * "); print(c, d); printf(" = "); res1 = a*c; res2 = b*d; print(res1, res2); printf("\n"); // case / print(a, b); printf(" / "); print(c, d); printf(" = "); if(c != 0) { if(c < 0) { d = -1*d; c = -1*c; } res1 = a*d; res2 = b*c; print(res1, res2); } else printf("Inf"); printf("\n"); } return 0; }