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// 题意:
// 输入两个整数N, H,按照字典序输出所有长度为N,恰好包含H个1的01串
//
规模:1<=H<=N<=16
// 算法A:2^N枚举,输出1的个数为H的。采用递归枚举
// 从bits[d]开始确定,已经用了c0个0和c1个1
#include<cstdio> #include<cstring> #include<iostream> #include<string> #include<algorithm> using namespace std; int n,h; int buf[16]; void solve(int c0, int c1, int d) { if(d==n) { if(c1==h) { for(int i=0;i<n;i++) printf("%d", buf[i]); printf("\n"); } return; } if(c0<n-h) { buf[d]=0; solve(c0+1, c1, d+1); } if(c1<h) { buf[d]=1; solve(c0, c1+1, d+1); } } int main() { int T; scanf("%d", &T); while(T--) { scanf("%d %d", &n, &h); solve(0, 0, 0); if(T) printf("\n"); } return 0; }
729 - The Hamming Distance Problem,码迷,mamicode.com
729 - The Hamming Distance Problem
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原文地址:http://www.cnblogs.com/cute/p/3696808.html