【BZOJ2752】【Luogu P2221】 [HAOI2012]高速公路

不是很难的一个题目。正确思路是统计每一条边被经过的次数，但我最初由于习惯直接先上了一个前缀和再推的式子，导致极其麻烦难以写对而且会爆\(longlong\)。

推导过程请看这里。

#include <bits/stdc++.h>
using namespace std;

const int N = 100000 + 5;

#define ls (p << 1)
#define rs (p << 1 | 1)
#define mid ((l + r) >> 1)
#define int long long

int n, m, sum1, sum2, sum3;

struct Segment_Tree {
    int tag[N << 2], s1[N << 2], s2[N << 2], s3[N << 2];
    
    Segment_Tree () {
        memset (s1, 0, sizeof (s1));
        memset (s2, 0, sizeof (s2));
        memset (s3, 0, sizeof (s3));
        memset (tag, 0, sizeof (tag));
    }
    
    void push_up (int p) {
        s1[p] = s1[ls] + s1[rs];
        s2[p] = s2[ls] + s2[rs];
        s3[p] = s3[ls] + s3[rs];
    }
    
    int F1 (int x, int y) {return y - x + 1;}  // \sum_{i = x} ^ {y} i ^ 0
    int F2 (int x, int y) {return (x + y) * (y - x + 1) / 2;} // \sum_{i = x} ^ {y} i ^ 1 
    int F3 (int x, int y) { 
        x = x - 1;
        int w1 = x * (x + 1) * (2 * x + 1) / 6;
        int w2 = y * (y + 1) * (2 * y + 1) / 6;
        return w2 - w1;
    } // \sum_{i = x} ^ {y} i ^ 2
    
    void work (int p, int l, int r, int val) {
        s1[p] += F1 (l, r) * val;
        s2[p] += F2 (l, r) * val;
        s3[p] += F3 (l, r) * val;
        tag[p] += val;
    }
    
    void push_down (int p, int l, int r) {
        work (ls, l, mid + 0, tag[p]);
        work (rs, mid + 1, r, tag[p]);
        tag[p] = 0;
    }
    
    void modify (int nl, int nr, int w, int l = 1, int r = n, int p = 1) {
        if (nl <= l && r <= nr) {
            work (p, l, r, w);
            return;
        }
        push_down (p, l, r);
        if (nl <= mid) modify (nl, nr, w, l, mid, ls);
        if (mid < nr) modify (nl, nr, w, mid + 1, r, rs);
        push_up (p); return;
    }
    
    void query (int nl, int nr, int l = 1, int r = n, int p = 1) {
        if (nl <= l && r <= nr) {
            sum1 += s1[p];
            sum2 += s2[p];
            sum3 += s3[p];
            return;
        }
        push_down (p, l, r);
        if (nl <= mid) query (nl, nr, l, mid, ls);
        if (mid < nr) query (nl, nr, mid + 1, r, rs);
        push_up (p); return;
    }
}tr; // 维护 \sum_{x = L}^{R} sumd(x) 

int gcd (int x, int y) {
    return y ? gcd (y, x % y) : x;
}

signed main () {
    freopen ("data.in", "r", stdin);
    cin >> n >> m;
    for (int i = 1; i <= m; ++i) {
        char opt; int l, r, v;
        cin >> opt;
        if (opt == 'C') {
            cin >> l >> r >> v; r--;
            tr.modify (l, r, v);
        } else {
            cin >> l >> r; r--;
            sum1 = sum2 = sum3 = 0;
            tr.query (l, r);
            int w1 = (r - l + 1 - r * l);
            int w2 = l + r;
            int w3 = -1;
            int upp = w1 * sum1 + w2 * sum2 + w3 * sum3;
            int dwn = (r - l + 2) * (r - l + 1) / 2;
            int d = gcd (upp, dwn); upp /= d, dwn /= d;
            cout << upp << "/" << dwn << endl;
        }
    }
}