hdu 5119 ^和>=m情况数+dp

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http://acm.hdu.edu.cn/showproblem.php?pid=5119

给一n元数列，求异或和>=m的情况数

dp[i][j]表示由前i个数组成异或和为j的方法数，

每次先获得不异或当前b[i]的情况数，加上异或了b[i]的情况数即可。
dp[0][0]=1,其他为0；

5s险过..

<span style="font-size:14px;">#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <queue>#include <map>#include <iostream>#include <sstream>#include <algorithm>using namespace std;#define RD(x) scanf("%d",&x)#define RD2(x,y) scanf("%d%d",&x,&y)#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define clr0(x) memset(x,0,sizeof(x))#define clr1(x) memset(x,-1,sizeof(x))#define eps 1e-9const double pi = acos(-1.0);typedef long long LL;const int inf = 1000000000;const int maxn = 2e6+5;int dp[45][maxn],n,m,b[45];LL work(){    RD2(n,m);    for(int i = 1;i <= n;++i)        RD(b[i]);    clr0(dp);    dp[0][0] = 1;    for(int i = 1;i <= n;++i){        for(int j = 0;j < maxn;++j)            dp[i][j] = dp[i-1][j];        for(int j = 0;j < maxn;++j)            dp[i][j^b[i]] += dp[i-1][j];//            dp[i][j] += dp[i-1][j] + dp[i-1][j^b[i]];    }    LL ans = 0;    for(int i = n;i <= n;++i)        for(int j = m;j < maxn;++j)            ans += dp[i][j];    return ans;}int main(){    int _,cas = 1;    RD(_);    while(_--){        printf("Case #%d: %I64d\n",cas++,work());    }    return 0;}/**/</span>

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