HDU5387-模拟水题

模拟钟表的时分秒针的走动，给出时间求出夹角。注意每组输出要有一个空格

以后要想好再写代码，这样一个水题做了50分钟，太弱了。。。

 #include<cstdio>
 #include<string>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 using namespace std;

 string s;
 int dp[][];
 int save[];

 int main(){
     int T;
     int cas = ;
     scanf("%d", &T);
     while(T--){
         cin >> s;
         memset(dp, , sizeof(dp));
         memset(save, , sizeof(save));
         for(int i = ; i < s.length(); i++){
             scanf("%d", &save[i]);
         }
         if(s[] == ''){
             dp[][] += save[];
             dp[][] = ;
         }else if(s[] == ''){
             dp[][] = dp[][] = ;
         }else{
             dp[][] += save[];
             dp[][] = ;
         }
         //printf("%d\n", max(dp[0][1], dp[0][0]));
         for(int i = ; i < s.length(); i++){
             if(s[i] == ''){
                 if(s[i-] == ''){
                     dp[i][] = dp[i-][] + save[i];
                 }else if(s[i-] == ''){
                     dp[i][] = dp[i-][];
                 }else if(s[i-] == '?'){
                     dp[i][] = max(dp[i-][] + save[i], dp[i-][]);
                 }
             }else if(s[i] == ''){
                 if(s[i-] == ''){
                     dp[i][] = dp[i-][];
                 }else if(s[i-] == ''){
                     dp[i][] = dp[i-][] + save[i];
                 }else if(s[i-] == '?'){
                     dp[i][] = max(dp[i-][] + save[i], dp[i-][]);
                     //printf("dp[%d][0] = %d \n", dp[i][0]);
                 }
             }else if(s[i] == '?'){
                 if(s[i-] == ''){
                     dp[i][] = dp[i-][] + save[i];
                     //printf("dp[%d][1] =  %d \n", i, dp[i][1]);
                     dp[i][] = dp[i-][];
                     //printf("dp[%d][0] = %d \n", i, dp[i][0]);
                 }else if(s[i-] == ''){
                     dp[i][] = dp[i-][] + save[i];
                     dp[i][] = dp[i-][];
                 }else if(s[i-] == '?'){
                     dp[i][] = max(dp[i-][] + save[i], dp[i-][]);
                     dp[i][] = max(dp[i-][] + save[i], dp[i-][]);
                 }
             }
             //printf("%d\n", max(dp[i][1], dp[i][0]));
         }
         int n = s.length() - ;
         printf("Case #%d: %d\n", ++cas, max(dp[n][], dp[n][]));
     }
     return ;
 }