题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5938

Four Operations

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1200    Accepted Submission(s): 352 Problem Description Little Ruins is a studious boy, recently he learned the four operations! Now he want to use four operations to generate a number, he takes a string which only contains digits  '1' -  '9', and split it into  5  intervals and add the four operations  '+',  '-',  '*' and  '/' in order, then calculate the result(/ used as integer division). Now please help him to get the largest result.   Input First line contains an integer  T , which indicates the number of test cases. Every test contains one line with a string only contains digits  '1'- '9'. Limits 1≤T≤105 5≤length of string≤20   Output For every test case, you should output  'Case #x: y', where  x indicates the case number and counts from  1 and  y is the result.   Sample Input 1 12345   Sample Output Case #1: 1   Source 2016年中国大学生程序设计竞赛（杭州）   Recommend liuyiding   Statistic |  Submit |  Discuss |  Note

题意：一个字符串，加上 + - * / 求最大数

解析：只有a+b-c*d/e得到的值才是最大值，那咱直接模拟减号的位置就好了，使a+b尽可能大，c*d/e尽可能小，那么c，d只有一位，a + b = max（a一位， b一位）;

代码：

#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<vector> #include<queue> #include<map> #include<cmath> #define N 109 using namespace std; const int INF = 0x3f3f3f3f; const int mod = 1e9 + 7; long long num[N][N]; char s[N]; void init() { for(int i = 1; s[i]; i++) { num[i][i - 1] = 0ll; for(int j = i; s[j]; j++) num[i][j] = num[i][j - 1] * 10ll + s[j] - '0'; } } int main() { int t, cnt = 0; scanf("%d", &t); while(t--) { scanf("%s", &s[1]); int len = strlen(&s[1]); init(); long long ans = -79; for(int i = 2; i <= len - 3; i++) ans = max(ans, max(num[1][1] + num[2][i], num[1][i - 1] + num[i][i]) - num[i + 1][i + 1] * num[i + 2][i + 2] / num[i + 3][len]); printf("Case #%d: %lld\n", ++cnt, ans); } return 0; }