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1 hdu 1032
[cpp] view plain copy print ? #include <iostream> using namespace std; int len(int n) { int s = 1; while (n!=1) { if (n&1) n = 3*n +1; else n /= 2; s++; } return s; } int main() { int i, j, n, l, m; while (~scanf("%d %d", &i, &j)) { printf("%d %d", i, j); if (i>j) swap(i, j); m = 0; for (;i<=j;++i) { l = len(i); if (m<l) m = l; } printf(" %d\n", m); } return 0; }2 hdu 1029 （鸽巢原理）
[cpp] view plain copy print ? #include <iostream> using namespace std; int main() { int ans, n, c, num; while (~scanf("%d", &n)) { c = 0; for (int i=1;i<=n;++i) { scanf("%d", &ans); if (c==0) { num = ans; c ++; } else if (ans == num) c ++; else c --; } printf("%d\n", num); } return 0; }3 hdu 1033 题目看到蛋碎
[cpp] view plain copy print ? #include <iostream> using namespace std; int f[4][2] = {0,10,10,0,0,-10,-10,0}; void move(int &x, int &y, char c, int &s) { int u; switch (s) { case 0: if (c=='V') u = 0, s = 1; else u = 2, s = 3; break; case 1: if (c=='V') u = 3, s = 2; else u = 1, s = 0; break; case 2: if (c=='V') u = 2, s = 3; else u = 0, s = 1; break; case 3: if (c=='V') u = 1, s = 0; else u = 3, s = 2; break; } x += f[u][0]; y += f[u][1]; } int main() { char c[300]; int i, k; while (~scanf("%s", c)) { k = strlen(c); printf("300 420 moveto\n310 420 lineto\n"); int x = 310, y = 420, s = 0; for (i=0;i<k;++i) { move(x, y, c[i], s); printf("%d %d lineto\n", x, y); } printf("stroke\nshowpage\n"); } return 0; }4 hdu 1036 sscanf的用法
[cpp] view plain copy print ? #include <iostream> using namespace std; int main() { int f, n, i, a, sum, h, m, s; double d; char c[20], t[100]; scanf("%d%lf", &n, &d); while (~scanf("%d", &a)) { sum = f = 0; for (i=1;i<=n;++i) { scanf("%s", c); if (c[0] == '-') f = 1; sscanf(c, "%d:%d:%d", &h, &m, &s); sum += h*3600 + m*60 + s; } if (f) printf("=: -\n", a); else { sum = int(sum/d + 0.5); printf("=: %d:%2.2d min/km\n", a, sum/60, sum`); } } return 0; }5 hdu 1037 不解释
[cpp] view plain copy print ? #include <iostream> using namespace std; int main() { int h1, h2, h3; while (~scanf("%d %d %d", &h1, &h2, &h3)) { if (h1>=168 && h2>=168 && h3>=168) printf("NO CRASH\n"); else { printf("CRASH "); if (h1<168) { printf("%d\n",h1); break; } if (h2<168) { printf("%d\n",h2); break; } if (h3<168) { printf("%d\n",h3); break; } } } return 0; }6 hdu 1038
[cpp] view plain copy print ? #include <stdio.h> #define P 3.1415926 int main() { double diameter, revolutions, time; double sum, zong; int n = 0; while(~scanf("%lf%lf%lf",&diameter,&revolutions,&time)) { if (revolutions==0) break; n++; sum = 0; zong = 0; sum = diameter / 12 / 5280 * P * revolutions * time; zong = sum/time; sum = zong / time * 60 * 60; printf("Trip #%d: %.2lf %.2lf\n",n,zong,sum); } return 0; }7 hdu 1039
[cpp] view plain copy print ? #include <stdio.h> #include <stdlib.h> #include <string.h> #include <assert.h> #define MAX_LEN 1111 char psd[MAX_LEN]; const char end[] = "end"; const char s1[] = "aeiou"; int c1(void) { int i,j; for(i=0;psd[i]!='\0';i++) for(j=0;s1[j]!='\0';j++) if(psd[i]==s1[j]) return 1; return 0; } int c2(void) { int i,j,cnt1,cnt2,tag; cnt1=cnt2=0; for(i=0;psd[i]!='\0';i++) { for(j=0,tag=0;s1[j]!='\0';j++) if(psd[i]==s1[j]) { cnt1++; cnt2=0; tag=1; break; } if(tag==0) { cnt1=0; cnt2++; } if(cnt1>=3 || cnt2>=3) return 0; } return 1; } int c3(void) { int i; for(i=1;psd[i]!='\0';i++) if(psd[i]==psd[i-1]) if(psd[i]!='e' && psd[i]!='o') return 0; return 1; } int main(void) { #ifndef ONLINE_JUDGE assert(freopen("1039.in","r",stdin)); #endif while(scanf("%s",psd),strcmp(psd,end)) { if(c1() && c2() && c3()) printf("<%s> is acceptable.\n",psd); else printf("<%s> is not acceptable.\n",psd); } return 0; }8 hdu 1047
大正整数加。
脑抽用string。。。越写越丑。。。还好一遍a了，避免了重敲的宿命。。
[cpp] view plain copy print ? #include <iostream> #include <string> #include <vector> #include <algorithm> #define clr(x, k) memset((x), (k), sizeof(x)) const int N = 1000; using namespace std; //char a[N], x[N]; string x, a; int l; void re(string &x) { int i = 0; int j = x.length() -1; while (i<j) swap(x[i++], x[j--]); } void add(string &a, string &x) { int i, j, ma, mi; ma = a.length(); mi = x.length(); if (ma<mi) { swap(ma, mi); swap(a, x); } for (i=0;i<mi;++i) a[i] += x[i]-'0'; for (i=0;i<ma;++i) if (a[i]>'9') { if (i+1<ma) a[i+1] += (a[i]-'0')/10; else a.insert(i+1,1,(a[i]-'0')/10+'0'); a[i] -= 10; } } int main() { int n; scanf("%d", &n); while (n--) { l = 1; a = "0"; while (cin>>x) { re(x); if (x == "0") { re(a); cout<<a<<endl; if (n!=0) cout<<endl; break; } add(a, x); } } return 0; }9 hdu 1856 ans初值为1，wa了几次才找到。。。
[cpp] view plain copy print ? #include <iostream> using namespace std; const int N = 10000002; int n, m; int f[N], r[N]; int find(int x) { if (f[x]!=x) f[x] = find(f[x]); return f[x]; } int main() { int x, y, i, ans; while (scanf("%d", &n)!=EOF) { for (i=0;i<=N;++i) { f[i] = i; r[i] = 1; } ans = 1; for (i=0;i<n;++i) { scanf("%d %d", &x, &y); x = find(x); y = find(y); if (x>y) swap(x, y); if (x!=y) { f[y] = x; r[x] += r[y]; if (r[x]>ans) ans = r[x]; } } printf("%d\n", ans); } return 0; }
10 hdu 1060
题目大意是输入N，求N^N的最高位数字。1<=N<=1,000,000,000 N^N = a*10^x; 我们要求的最右边的数字就是(int)a，即a的整数部分 两边同时取以10为底的对数 lg(N^N) = lg(a*10^x) ; N*lg(N) = lg(a) + x;
N*lg(N) - x = lg(a)
a = 10^(N*lg(N) - x);
现在就只有x是未知的了，如果能用n来表示x的话，这题就解出来了。
又因为，x是N^N的位数-1。比如 N^N = 1200 ==> x = 3;
实际上就是 x 就是lg(N^N) 向下取整数，表示为[lg(N^N)]
a = 10^(N*lg(N) – [lg(N^N)]);
然后(int)a 就是答案了。
[cpp] view plain copy print ? #include<iostream> #include<cstring> #include<cmath> using namespace std; int main(){ int t; __int64 ans,n; double m; scanf("%d",&t); while(t--){ scanf("%I64d",&n); m=n*log10(n+0.0); m-=(__int64)m; ans=pow((double)10,m); printf("%I64d\n",ans); } return 0; }11 hdu 1061 N^N
[cpp] view plain copy print ? #include <cstdio> #include <cstdlib> #define I64 __int64 int Fuction(I64 n) { int res = 1; I64 b = n; if(!n) // n==0, 输出1 return 1; if(!(n % 10)) return 0; while(b) { if(b & 1) { res *= n; res %= 10; // 要取模，否则溢出 } n *= n; n %= 10; // 要取模，否则溢出 b >>= 1; } return res % 10; } int main() { int t; I64 n; scanf("%d", &t); while(t--) { scanf("%I64d", &n); printf("%d\n", Fuction(n)); } return 0; } //题目分析：求n^n的个位数，只要根据每一个数的幂的周期性规律，就行了 #include <stdio.h> int main() { int m,n,a; int s[10][4]={{0},{1},{6,2,4,8},{1,3,9,7},{6,4},{5},{6}, {1,7,9,3},{6,8,4,2},{1,9}}; scanf("%d",&m); while(m--) { scanf("%d",&n); a=n % 10; if(a==0||a==1||a==5||a==6) printf("%d\n",a); else if(a==4||a==9) printf("%d\n",s[a][n%2]); else if(a==2||a==3||a==7||a==8) printf("%d\n",s[a][n%4]); } return 0; }