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Re:因为是每次都找最小的一个,直到三个数都相等,所以应该没有错呀,总不能出现负数吧

Posted by hahaxiao at 2005-09-17 10:21:22 on Problem 1006
In Reply To:why wrong answer?  Posted by:hahaxiao at 2005-09-16 17:49:28
> //每次找三个数中最小的一个,然后加上相应有23 28 33,直到三个数相等
> //判断如果这三个数比第四个数小或大两种情况,
> #include <iostream>
> using namespace std;
> int main() {
>     int a[4];
>     int k,i,p;
>     int min;
>     k=0;
>     while(1){ 
>         cin>>a[0]>>a[1]>>a[2]>>a[3];
>         if(a[0]==-1 && a[1]==-1 && a[2]==-1 && a[3]==-1) break;
>         while(a[0]!=a[1] || a[1]!=a[2] ) {
>             min=32767;
>             for(i=0;i<3;i++){
>               if(a[i]<min) {
>                 min=a[i];
>                 p=i;
>               }
>             }
>             if(p==0) a[p]=a[p]+23;
>             if(p==1) a[p]=a[p]+28;
>             if(p==2) a[p]=a[p]+33;
>         }
>         ++k;
>         if(a[0]<=a[3]) cout<<"Case "<<k<<": the next triple peak occurs in "<<21252+a[0]-a[3]<<" days."<<endl;
>         else cout<<"Case "<<k<<": the next triple peak occurs in "<<a[0]-a[3]<<" days."<<endl;
>     }
>     return 0;
> }        
>     
>                    
>         

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