【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;
		}
	}
}
posted @ 2019-05-28 20:15  maomao9173  阅读(82)  评论(0编辑  收藏  举报