BZOJ3527 推出卷积公式FFT求值

传送门：https://www.lydsy.com/JudgeOnline/problem.php?id=3527

题意：

\(F_{j}=\sum_{i<j} \frac{q_{i} q_{j}}{(i-j)^{2}}-\sum_{i>j} \frac{q_{i} q_{j}}{(i-j)^{2}}\)

求\(E_i=F_i/q_i\)

题解：

推公式：

\[E_i=F_i/q_i\\
E_i=\sum_{j=i}^{n}\frac{q_j}{(i-j)^2}-\sum_{j=0}^{i}\frac{q_j}{(i-j)^2}\\
设函数f(i)为q_i,g(i)为(i)^2\\
\sum_{i=0}^{j} f_{i} * g_{j-i}\\
\begin{array}{l}{\sum_{i>j} \frac{q_{i}}{(i-j)^{2}}=\sum_{i=j}^{n} \frac{q_{i}}{(i-j)^{2}}} {=\sum_{i=0}^{n-j} \frac{q_{n-i}}{(j-i)^{2}}=\sum_{i=0}^{n-j} f_{n-i} * g_{i-j}}\end{array}
\]

于是我们求出f和g 后fft，然后求值即可

代码：

#include <set>

#include <map>

#include <cmath>

#include <cstdio>

#include <string>

#include <vector>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

typedef long long LL;

typedef pair<int, int> pii;

typedef unsigned long long uLL;

#define ls rt<<1

#define rs rt<<1|1

#define lson l,mid,rt<<1

#define rson mid+1,r,rt<<1|1

#define bug printf("*********\n")

#define FIN freopen("input.txt","r",stdin);

#define FON freopen("output.txt","w+",stdout);

#define IO ios::sync_with_stdio(false),cin.tie(0)

#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"

#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"

#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"

const int maxn = 1e6 + 5;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + 7;

const double Pi = acos(-1.0);

LL quick_pow(LL x, LL y) {

    LL ans = 1;

    while(y) {

        if(y & 1) {

            ans = ans * x % mod;

        } x = x * x % mod;

        y >>= 1;

    } return ans;

}

struct complex {

    double x, y;

    complex(double xx = 0, double yy = 0) {

        x = xx, y = yy;

    }

} f[maxn], f1[maxn], g[maxn];

complex operator + (complex a, complex b) {

    return complex(a.x + b.x, a.y + b.y);

}

complex operator - (complex a, complex b) {

    return complex(a.x - b.x, a.y - b.y);

}

complex operator * (complex a, complex b) {

    return complex(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);

}

int n, m;

int l, r[maxn];

int limit = 1;

void fft(complex *A, int type) {

    for(int i = 0; i < limit; i++) {

        if(i < r[i]) swap(A[i], A[r[i]]);

    }

    for(int mid = 1; mid < limit; mid <<= 1) {

        complex Wn(cos(Pi / mid), type * sin(Pi / mid));

        for(int R = mid << 1, j = 0; j < limit; j += R) {

            complex w(1, 0);

            for(int k = 0; k < mid; k++, w = w * Wn) {

                complex x = A[j + k], y = w * A[j + mid + k];

                A[j + k] = x + y;

                A[j + k + mid] = x - y;

            }

        }

    }

}

int ans[maxn];

char numA[maxn], numB[maxn];

int main() {

#ifndef ONLINE_JUDGE

    FIN

#endif

    int n;

    while(scanf("%d", &n) != EOF) {

        n--;

        m = n * 2;

        for(int i = 0; i <= n; i++) {

            scanf("%lf", &f[i].x);

            // debug1(f[i].x);

            f1[n - i].x = f[i].x;

        }

        // bug;

        for(int i = 1; i <= n; i++) {

            g[i].x = (double)(1.0 / i / i);

        }

        while(limit <= m) limit <<= 1, l++;

        for(int i = 0; i <= limit; i++) {

            r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));

        }

        fft(f, 1);

        fft(f1, 1);

        fft(g, 1);

        for(int i = 0; i <= limit; i++) {

            f[i] = f[i] * g[i];

        }

        for(int i = 0; i <= limit; i++) {

            g[i] = f1[i] * g[i];

        }

        fft(f, -1);

        fft(g, -1);

        for(int i = 0; i <= limit; i++) {

            f[i].x = f[i].x / limit;

        }

        for(int i = 0; i <= limit; i++) {

            g[i].x = g[i].x / limit;

        }

        int t = m / 2;

        for(int i = 0; i <= t; i++) {

            printf("%.3f\n", f[i].x - g[n - i].x);

        }

    }

    return 0;

}

巴特西

BZOJ3527 推出卷积公式FFT求值

BZOJ3527 推出卷积公式FFT求值

题意：

题解：

代码：

最新文章

热门文章