BZOJ4589: Hard Nim(FWT 快速幂)

题意

题目链接

sol

神仙题orzzzz

题目可以转化为从\(\leqslant m\)的质数中选出\(n\)个\(xor\)和为\(0\)的方案数

这样就好做多了

设\(f(x) = [x \text{是质数}]\)

\(n\)次异或fwt即可

快速幂优化一下，中间不用ifwt，最后转一次就行(然而并不知道为什么)

哪位大佬教教我这题的dp怎么写呀qwqqqq

死过不过去样例。。

#include<bits/stdc++.h>
using namespace std;
const int maxn = (1 << 17) + 10, mod = 998244353, inv2 = 499122177;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, a[maxn], b[maxn], c[maxn];
int add(int x, int y) {
    if(x + y < 0) return x + y + mod;
    return x + y >= mod ? x + y - mod : x + y;
}
int mul(int x, int y) {
    return 1ll * x * y % mod;
}
void fwtor(int *a, int opt) {
    for(int mid = 1; mid < n; mid <<= 1) 
        for(int r = mid << 1, j = 0; j < n; j += r)
            for(int k = 0; k < mid; k++) 
                if(opt == 1) a[j + k + mid] = add(a[j + k], a[j + k + mid]);
                else a[j + k + mid] = add(a[j + k + mid], -a[j + k]);
}
void fwtand(int *a, int opt) {
    for(int mid = 1; mid < n; mid <<= 1) 
        for(int r = mid << 1, j = 0; j < n; j += r)
            for(int k = 0; k < mid; k++) 
                if(opt == 1) a[j + k] = add(a[j + k], a[j + k + mid]);
                else a[j + k] = add(a[j + k], -a[j + k + mid]);
}
void fwtxor(int *a, int opt) {
    for(int mid = 1; mid < n; mid <<= 1) 
        for(int r = mid << 1, j = 0; j < n; j += r)
            for(int k = 0; k < mid; k++) {
                int x = a[j + k], y = a[j + k + mid];
                if(opt == 1) a[j + k] = add(x, y), a[j + k + mid] = add(x, -y);
                else a[j + k] = mul(add(x, y), inv2), a[j + k + mid] = mul(add(x, -y), inv2);               
            }

}
int main() {
    n = 1 << (read());
    for(int i = 0; i < n; i++) a[i] = read();
    for(int i = 0; i < n; i++) b[i] = read();
    fwtor(a, 1); fwtor(b, 1);
    for(int i = 0; i < n; i++) c[i] = mul(a[i], b[i]);
    fwtor(c, -1); fwtor(a, -1); fwtor(b, -1);
    for(int i = 0; i < n; i++) printf("%d ", c[i]); puts("");
    fwtand(a, 1); fwtand(b, 1);
    for(int i = 0; i < n; i++) c[i] = mul(a[i], b[i]);
    fwtand(c, -1); fwtand(a, -1); fwtand(b, -1);    
    for(int i = 0; i < n; i++) printf("%d ", c[i]); puts("");
    fwtxor(a, 1); fwtxor(b, 1);
    for(int i = 0; i < n; i++) c[i] = mul(a[i], b[i]);
    fwtxor(c, -1); fwtxor(a, -1); fwtxor(b, -1);    
    for(int i = 0; i < n; i++) printf("%d ", c[i]);
    return 0;
}