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哪位帮我看看代码啊,我觉得没问题,可就是 WA#include<iostream> #include<math.h> #define N 3000 using namespace std; int main() { int n,i,j; int a[N],index[N]; while(cin>>n) { if(n==1) { cout<<"Jolly"<<endl; continue; } else { i=0; while(i<n) { cin>>a[i]; index[i]=0; ++i; } i=0; while(i<n-1) { j=abs(a[i]-a[i+1]); if(j>=n||index[j]==1) { cout<<"Not jolly"<<endl; break; } index[j]=1; ++i; } if(i==n-1)cout<<"Jolly"<<endl; } } return 0; } 我的想法很简单,用一个指标数组来指示1到n-1的数哪些已经用过了,如果重复出现,则不是;如果能执行到最后,则是 Followed by:
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