CF993E Nikita and Order Statistics 【fft】

最新推荐文章于 2021-10-26 21:02:15 发布

weixin_30713953

最新推荐文章于 2021-10-26 21:02:15 发布

阅读量99

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/Mychael/p/9285521.html

本文提供CF993E题目的详细解答方案，利用前缀和与卷积技巧解决区间统计问题，通过FFT算法优化计算过程，实现高效求解。

题目链接

CF993E

题解

我们记小于\(x\)的位置为\(1\)，否则为\(0\)
区间由端点决定，转为两点前缀和相减
我们统计出每一种前缀和个数，记为\(A[i]\)表示值为\(i\)的位置出现的次数
那么对于\(k > 0\)有
\[ans_k = \sum\limits_{x - y = k} A[x]A[y]\]
令
\[B[x] = A[n - x]\]
那么有
\[ans_k = \sum\limits_{x + y = n + k} A[x]B[y]\]
就成了卷积的形式
第\(n + k\)项系数就是\(ans_k \qquad k > 0\)

对于\(k = 0\)，可以直接统计，也可以减去卷积中重复的部分
首先减去空串的个数\(n + 1\)，然后再除以\(2\)【因为当\(x\)和\(y\)相等，大小顺序就可以颠倒了】
最后求得的就是\(k = 0\)的答案

复杂度\(O(nlogn)\)

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 800005,maxm = 100005,INF = 0x3f3f3f3f;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
    return flag ? out : -out;
}
struct E{
    double a,b;
    E(){}
    E(double x,double y):a(x),b(y) {}
    E(int x,int y):a(x),b(y) {}
    inline E operator =(const int& b){
        this->a = b; this->b = 0;
        return *this;
    }
    inline E operator =(const double& b){
        this->a = b; this->b = 0;
        return *this;
    }
    inline E operator /=(const double& b){
        this->a /= b; this->b /= b;
        return *this;
    }
};
inline E operator *(const E& a,const E& b){
    return E(a.a * b.a - a.b * b.b,a.a * b.b + a.b * b.a);
}
inline E operator *=(E& a,const E& b){
    return a = E(a.a * b.a - a.b * b.b,a.a * b.b + a.b * b.a);
}
inline E operator +(const E& a,const E& b){
    return E(a.a + b.a,a.b + b.b);
}
inline E operator -(const E& a,const E& b){
    return E(a.a - b.a,a.b - b.b);
}
const double pi = acos(-1);
int R[maxn];
void fft(E* a,int n,int f){
    for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
    for (int i = 1; i < n; i <<= 1){
        E wn(cos(pi / i),f * sin(pi / i));
        for (int j = 0; j < n; j += (i << 1)){
            E w(1,0),x,y;
            for (int k = 0; k < i; k++,w = w * wn){
                x = a[j + k],y = w * a[j + k + i];
                a[j + k] = x + y; a[j + k + i] = x - y;
            }
        }
    }
    if (f == -1) for (int i = 0; i < n; i++) a[i] /= n;
}
E A[maxn],B[maxn];
int cnt[maxn],N,x,a[maxn];
int main(){
    N = read(); x = read();
    cnt[0]++;
    REP(i,N) cnt[a[i] = a[i - 1] + (read() < x ? 1 : 0)]++;
    for (int i = 0; i <= N; i++) A[i] = cnt[i],B[i] = cnt[N - i];
    int n = 1,L = 0;
    while (n <= (N << 1)) n <<= 1,L++;
    for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    fft(A,n,1); fft(B,n,1);
    for (int i = 0; i < n; i++) A[i] *= B[i];
    fft(A,n,-1);
    A[N].a -= N + 1; A[N].a /= 2;
    for (int i = 0; i <= N; i++)
        printf("%.0lf ",floor(A[N + i].a + 0.3));
    return 0;
}

转载于:https://www.cnblogs.com/Mychael/p/9285521.html