标签:style blog http io ar color os sp for
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5119
分析:dp[i][j]表示由前i个数组成异或和为j的方法数,则dp[i][j]=d[i-1][j^a[i]]+dp[i][j];
边界:dp[0][0]=1,其他为0;复杂度40*1e6
#include <cstdio> #include <cstring> #include <cmath> #include <iostream> #include <algorithm> #include <queue> #include <cstdlib> #include <vector> #include <set> #include <map> #define LL long long #define inf 1<<30 #define mod 1000000007 using namespace std; int a[100]; int dp[2][(1<<20)]; int main() { int T,n,m; int cas=1; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); memset(dp,0,sizeof(dp)); for(int i=1; i<=n; i++)scanf("%d",&a[i]); int pre=0,cur=1; dp[0][0]=1; for(int i=1; i<=n; i++) { memcpy(dp[cur],dp[pre],sizeof(dp[pre]));//小于i个数异或和为j的方法要赋值给现在 for(int j=0; j<=(1<<20)-1; j++) { int num=j^a[i]; if(dp[pre][num]) { dp[cur][j]+=dp[pre][num]; } } swap(pre,cur); } LL ans=0; for(int i=m; i<=(1<<20)-1; i++)ans+=dp[pre][i]; printf("Case #%d: ",cas++); printf("%I64d\n",ans); } }
标签:style blog http io ar color os sp for
原文地址:http://www.cnblogs.com/lienus/p/4131678.html