BZOJ4589: Hard Nim(FWT 快速冪)
阿新 • • 發佈:2018-11-29
oid printf get .org xor etc i++ ref har
題意
題目鏈接
Sol
神仙題Orzzzz
題目可以轉化為從\(\leqslant M\)的質數中選出\(N\)個\(xor\)和為\(0\)的方案數
這樣就好做多了
設\(f(x) = [x \text{是質數}]\)
\(n\)次異或FWT即可
快速冪優化一下,中間不用IFWT,最後轉一次就行(然而並不知道為什麽)
哪位大佬教教我這題的DP怎麽寫呀qwqqqq
死過不過去樣例。。
#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; }
BZOJ4589: Hard Nim(FWT 快速冪)