Integer Game (UVA11489) 3 multiple, integeruva11489

Integer Game (UVA11489) 3 multiple, integeruva11489
K-Integer Game

Time Limit:1000 MSMemory Limit:0 KB64bit IO Format:% Lld & % llu

Submit Status Practice ultraviolet A 11489

 

 

 

 

In n, the number is a multiple of 3. If not, the result fails. Thought: if you can make the current number a multiple of 3, the number and must be a multiple of 3. To maintain this state, the remainder of each bitwise is obtained first, and the count of cnt [0], cnt [1], cnt [2]; is calculated, whether or not the entire string can be divisible by 3 depends on ans = (cnt [1] + cnt [2] * 2) % 3; the situation is discussed. If the entire string can be divisible, determine the number of cnt [0]. If it is an even number, T; otherwise, S; if it cannot be divisible, determine whether the number of cnt [ans] is greater than 0; <eg: 11111>; if the value is greater than 0, the number of cnt [0] is an even number. Otherwise, T; if the value is equal to 0, T; always in the 11111 condition! O (︶ ︿ ︶) o alas reprinted please indicate the source: Search for &

 

 1 #include<stdio.h> 2 #include<string.h> 3 int main() 4 { 5     int T,len,ca=1; 6     scanf("%d",&T); 7     char a[10005]; 8  9     while(ca<=T)10     {11         scanf("%s",a);12         len=strlen(a);13         int cnt[3]={0,0,0};14         for(int i=0;i<len;i++)15         {16             int tp=a[i]-'0';17             cnt[tp%3]++;18         }19         printf("Case %d: ",ca++);20         int mo=(cnt[1]+cnt[2]*2)%3;21         if(mo)22         {23             if(cnt[mo]>0)//1111124             {25                 if(cnt[0]&1) printf("T\n");26                 else printf("S\n");27             }28             else29             {30                 printf("T\n");31             }32         }33         else34         {35             if(cnt[0]&1) printf("S\n");36             else printf("T\n");37         }38     }39     return 0;40 }