BZOJ4589: Hard Nim

FWT模板题。

#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define MOD 1000000007
#define rep(i,x,y) for(int i=(x);i<=(y);++i)
#define drep(i,x,y) for(int i=(x);i>=(y);--i)
inline int read(){
    char ch=getchar();int x=0,f=1;
    while(ch<'0'||ch>'9'){
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while('0'<=ch&&ch<='9'){
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
inline int Power(int x, int y) {
    int ret = 1;
    while(y) {
        if(y & 1) ret = 1ll * ret * x % MOD;
        x = 1ll * x * x % MOD; y >>= 1;
    }
    return ret;
}
inline void FWT_xor(int *a, int N, int op) {
    int inv2 = (MOD + 1) / 2;
    for(int d = 1; d < N; d <<= 1) {
        for(int i = 0; i < N; i += (d << 1)) {
            for(int j = 0; j < d; ++ j) {
                int X = a[i + j], Y = a[i + j + d];
                a[i + j] = (X + Y) % MOD;
                a[i + j + d] = (X + MOD - Y) % MOD;
                if(op == -1) {
                    a[i + j] = 1ll * a[i + j] * inv2 % MOD;
                    a[i + j + d] = 1ll * a[i + j + d] * inv2 % MOD;
                }
            }
        }
    }
}
int a[200010];
int pri[200010], tot;
bool vis[200010];
inline void init() {
    for(int i = 2; i <= 50000; ++ i) {
        if(!vis[i]) {
            pri[++ tot] = i;
            for(int j = i + i; j <= 50000; j += i) {
                vis[j] = 1;
            }
        }
    }
}
int main(){
    int n, m;
    init();
    while(scanf("%d%d", &n, &m) != EOF) {
        int N;
        for(N = 1; N <= m; N <<= 1);
        for(int i = 0; i < N; ++ i) a[i] = 0;
        for(int i = 1; i <= tot; ++ i) {
            if(pri[i] > m) break;
            a[pri[i]] = 1;
        }
        FWT_xor(a, N, 1);
        for(int i = 0; i < N; ++ i) {
            a[i] = Power(a[i], n);
        }
        FWT_xor(a, N, -1);
        printf("%d
", a[0]);
    }
}