乱搞:
rt。有1就能输出全部的数,否则仅仅能输出偶数
Answers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 489 Accepted Submission(s): 294
Problem Description
Sample Input
2 4 2 2 2 1 0 1 2 3
Sample Output
NO YES YES YES
Source
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int n,m; int t[100100],c[100100]; int main() { while(scanf("%d%d",&n,&m)!=EOF) { bool flag=false; for(int i=1;i<=n;i++) scanf("%d",t+i); for(int i=1;i<=n;i++) { scanf("%d",c+i); if(c[i]==1) flag=true; } for(int i=0;i<m;i++) { int x; scanf("%d",&x); if(x<=0) puts("NO"); else { if(flag) puts("YES"); else { if(x%2==0) puts("YES"); else puts("NO"); } } } } return 0; }