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  • 洛谷P4717 【模板】快速沃尔什变换(FWT)

    题意

    题目链接

    Sol

    背板子背板子

    #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;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/10040419.html
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