usaco training-Hamming Codes

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//usaco training-Hamming Codes/* * 题目类型：搜索 * N<64 ，B<=8由于8位二进制数共有511个，数据量比较小可以直接暴力搜索 * res数组记录结果值，逐个搜索当前值，加入到res中 * 每次搜索从res最后的数字的下一个数开始，检查是否满足条件 * ！！！注意输出，N=10时，注意不要多输出空行 *//*Compiling...Compile: OKExecuting... Test 1: TEST OK [0.000 secs, 3028 KB] Test 2: TEST OK [0.000 secs, 3028 KB] Test 3: TEST OK [0.000 secs, 3028 KB] Test 4: TEST OK [0.000 secs, 3028 KB] Test 5: TEST OK [0.000 secs, 3028 KB] Test 6: TEST OK [0.000 secs, 3028 KB] Test 7: TEST OK [0.000 secs, 3028 KB] Test 8: TEST OK [0.000 secs, 3028 KB] Test 9: TEST OK [0.000 secs, 3028 KB] Test 10: TEST OK [0.000 secs, 3028 KB] Test 11: TEST OK [0.000 secs, 3028 KB]All tests OK. */#include <iostream>#include <cmath>#include <fstream>using namespace std;int res[70];int N,B,D;bool dist(int a,int b){int i,num=0;for(i=1; i<=B; ++i){if((a&1)!=(b&1))num++;a=a>>1; b=b>>1;}return (num>=D) ? true : false; }int main(void){int i,j,tmp;ifstream fin("hamming.in");ofstream fout("hamming.out");fin>>N>>B>>D;for(i=1; i<N; ++i)for(tmp=res[i-1]+1; tmp<(int)pow((double)2,B); ++tmp){for(j=0; j<i; ++j)if(!dist(tmp,res[j]))break;if(j==i){res[i]=tmp;break;}}for(i=0; i!=N; ++i){if(!((i+1)%10))fout<<res[i]<<endl;else if(i!=N-1)fout<<res[i]<<' ';elsefout<<res[i];}if(N%10)fout<<endl;return 0;}