Codeforces Round #438 C Qualification Rounds(暴力+位运算)
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题目链接:点击打开链接
因为k最大为4,所以可以根据二进制转十进制的策略存储所有的题目种类
只需要根据k的值找到一对数,使其表示成二进制之后每个数位上1最多出现一次
暴力一遍即可
AC代码如下:
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int n, k;int problem[16];int trans[] = {1,2,4,8,16};inline int getBit(int n, int p){ int temp = n>>p; return temp%2;}int solve(int m){ if(problem[0]) return 1; for(int i=1;i<=m;++i) { if(problem[i]) { for(int j=i+1;j<=m;++j) { if(problem[j]) { int flag=1; for(int t=0;t<k;++t) if(getBit(i,t)&&getBit(j,t)) { flag=0; break; } if(flag) return 1; } } } } return 0;}int main(){ ios::sync_with_stdio(0); cin.tie(0); while(cin>>n>>k) { memset(problem,0,sizeof(problem)); int temp; for(int i=1;i<=n;++i) { temp = 0; for(int j=0;j<k;++j) { int a; cin>>a; temp += trans[j]*a; } problem[temp]++; } if(solve(trans[k]-1)) cout<<"YES"<<endl; else cout<<"NO"<<endl; } return 0;}
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