4821: [Sdoi2017]相关分析

4821: [Sdoi2017]相关分析

链接

分析：

　　大力拆式子，化简，然后线段树。注意精度问题与爆longlong问题。

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
#define Root 1, n, 1
#define lc rt << 1
#define rc rt << 1 | 1
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
using namespace std;
typedef double DB;

inline int read() {
    int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
    for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
}

const int N = ;
DB sx[N << ], sy[N << ], s[N << ], s2[N << ], tag[N << ], taga[N << ], tagb[N << ], X[N], Y[N];
DB sum[N], sum2[N];
DB Sx, Sy, S, S2;
int n;

inline void pushup(int rt) {
    sx[rt] = sx[lc] + sx[rc];
    sy[rt] = sy[lc] + sy[rc];
    s2[rt] = s2[lc] + s2[rc];
    s[rt] = s[lc] + s[rc];
}
inline void col(int rt,int l,int r) {
    sy[rt] = sx[rt] = sum[r] - sum[l - ];
    s2[rt] = s[rt] = sum2[r] - sum2[l - ];
    tag[rt] = ; taga[rt] = tagb[rt] = ;
}
inline void add(int rt,DB len,DB a,DB b) {
    s[rt] += sx[rt] * b + sy[rt] * a + a * b * len;
    s2[rt] += a * a * len +  * a * sx[rt];
    sx[rt] += a * len;
    sy[rt] += b * len;
    taga[rt] += a, tagb[rt] += b;
}
inline void pushdown(int rt,int l,int r) {
    int mid = (l + r) >> ;
    if (tag[rt]) {
        col(lc, l, mid); col(rc, mid + , r);
        tag[rt] = ;
    }
    if (taga[rt] || tagb[rt]) { // !!!
        add(lc, mid - l + , taga[rt], tagb[rt]);
        add(rc, r - mid, taga[rt], tagb[rt]);
        taga[rt] = tagb[rt] = ;
    }
}
void build(int l,int r,int rt) {
    if (l == r) {
        sx[rt] = X[l], sy[rt] = Y[l], s[rt] = X[l] * Y[l], s2[rt] = X[l] * X[l]; return ;
    }
    int mid = (l + r) >> ;
    build(lson), build(rson);
    pushup(rt);
}
void update(int l,int r,int rt,int L,int R,DB a,DB b) {
    if (L <= l && r <= R) {
        add(rt, r - l + , a, b); return ;
    }
    int mid = (l + r) >> ;
    pushdown(rt, l, r);
    if (L <= mid) update(lson, L, R, a, b);
    if (R > mid) update(rson, L, R, a, b);
    pushup(rt);
}
void Change(int l,int r,int rt,int L,int R) {
    if (L <= l && r <= R) {
        col(rt, l, r); return ;
    }
    int mid = (l + r) >> ;
    pushdown(rt, l, r);
    if (L <= mid) Change(lson, L, R);
    if (R > mid) Change(rson, L, R);
    pushup(rt);
}
void query(int l,int r,int rt,int L,int R) {
    if (L <= l && r <= R) {
        Sx += sx[rt], Sy += sy[rt], S += s[rt], S2 += s2[rt]; return ;
    }
    int mid = (l + r) >> ;
    pushdown(rt, l, r);
    if (L <= mid) query(lson, L, R);
    if (R > mid) query(rson, L, R);
}
void Ask() {
    int l = read(), r = read();
    Sx = Sy = S = S2 = ;
    query(Root, l, r);
    double x = 1.0 * Sx / (r - l + ), y = 1.0 * Sy / (r - l + );
//    double u = S - Sy * x - Sx * y + x * y * (r - l + 1);
//    double d = S2 + x * x * (r - l + 1) - 2 * Sx * x;
    double u = S - y * Sx, d = S2 - x * Sx;
    printf("%.10lf\n", (double)(u / d));
}
void work1() {
    int l = read(), r = read();
    DB a = (DB)read(), b = (DB)read();
    update(Root, l, r, a, b);
}
void work2() {
    int l = read(), r = read();
    DB a = (DB)read(), b = (DB)read();
    Change(Root, l, r);
    update(Root, l, r, a, b);
}
int main() {
    n = read();int m = read();
    for (int i = ; i <= n; ++i) {
        sum[i] = sum[i - ] + 1.0 * i;
        sum2[i] = sum2[i - ] + 1.0 * i * i; // 此处报int !!!
    }
    for (int i = ; i <= n; ++i) X[i] = (DB)read();
    for (int i = ; i <= n; ++i) Y[i] = (DB)read();
    build(Root);
    while (m --) {
        int opt = read();
        if (opt == ) Ask();
        else if (opt == ) work1();
        else work2();
    }
    return ;
}