HDU-6266 发表于 2018-04-14 | 分类于 ACM | 浏览 次 总结要合理分析题目难度,不要高估,检查时要检查所有情况,不要只盯着一个地方。 代码123456789101112131415161718192021222324252627282930313233343536373839404142434445464748#include "stdio.h"#include "cstring"#include "stdlib.h"#include "map"#include "string"#include "iostream"#include "set"#include "queue"#include "vector"#define FOP freopen("input.txt","r",stdin)#define Met(x,y) memset(x,y,sizeof x)#define L index*2#define R index*2+1using namespace std;typedef long long ll;const ll MAXN=1e6+100;const ll MOD=1e9+7;ll a[MAXN];int main(int argc, char const *argv[]) { // cout<<188*0.6<<endl; ll T_T; cin>>T_T; while(T_T--) { ll n,d; cin>>n>>d; bool a1=1; ll n1=0; for(int i=0;i<n;++i) { scanf("%lld",&a[i]); if(a[i]==1) n1++; // printf("n1==%lld\n",n1); } if (d==1) { if (n1==n && n%3==0) { puts("No"); } else { puts("Yes"); } } else { if((n1==n-1 && n%3==0) || (n1==n && (n-1)%3==0) ||(n1==n-1 && (n-1)%3==0)) puts("No"); else puts("Yes"); } }}