? ? ? ? ? ? ? ? ? ? ? ?Happy Matt Friends Matt has N friends. They are playing a game together.?

Each of Matt’s friends has a magic number. In the game, Matt selects some (could be zero) of his friends. If the xor (exclusive-or) sum of the selected friends’magic numbers is no less than M , Matt wins.?

Matt wants to know the number of ways to win.

InputThe first line contains only one integer T , which indicates the number of test cases.?

For each test case, the first line contains two integers N, M (1 ≤ N ≤ 40, 0 ≤ M ≤ 10?⁶).?

In the second line, there are N integers ki (0 ≤ k?_i?≤ 10?⁶), indicating the i-th friend’s magic number.OutputFor each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y indicates the number of ways where Matt can win.Sample Input

2 3 2 1 2 3 3 3 1 2 3

Sample Output

Case #1: 4 Case #2: 2

Hint

In the ?rst sample, Matt can win by selecting: friend with number 1 and friend with number 2. The xor sum is 3. friend with number 1 and friend with number 3. The xor sum is 2. friend with number 2. The xor sum is 2. friend with number 3. The xor sum is 3. Hence, the answer is 4.


14年北京站題~
題目意思：給出t組數(shù)據(jù) n個數(shù) 求XOR出大于m的數(shù)的個數(shù)，每個數(shù)都有兩種狀態(tài)，取與不取。其實(shí)就是一個01背包。 #include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e6+10; const int MOD=1e9+7; ll a[maxn]; int dp[45][maxn],n,m; ll slove() {dp[1][0]=dp[1][a[1]]=1;for(int i=2; i<=n; i++){for(ll j=0; j<maxn; j++){dp[i][j]+=dp[i-1][j];dp[i][j^a[i]]+=dp[i-1][j];}}ll ans=0;for(ll i=m; i<maxn; i++){ans+=dp[n][i];}return ans; } int main() {int t,k=0;scanf("%d",&t);while(t--){memset(dp,0,sizeof(dp));scanf("%d %d",&n,&m);for(int i=1; i<=n; i++)scanf("%lld",&a[i]);ll ans=slove();printf("Case #%d: %lld\n",++k,ans);}return 0; }

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PS：摸魚怪的博客分享，歡迎感謝各路大牛的指點(diǎn)~

轉(zhuǎn)載于:https://www.cnblogs.com/MengX/p/9074603.html

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