徐州网络赛H-Ryuji doesn't want to study【线段树】

Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i]a[i].

Unfortunately, the longer he learns, the fewer he gets.

That means, if he reads books from ll to rr, he will get a[l] \times L + a[l+1] \times (L-1) + \cdots + a[r-1] \times 2 + a[r]a[l]×L+a[l+1]×(L−1)+⋯+a[r−1]×2+a[r](LL is the length of [ ll, rr ] that equals to r - l + 1r−l+1).

Now Ryuji has qq questions, you should answer him:

11. If the question type is 11, you should answer how much knowledge he will get after he reads books [ ll, rr ].

22. If the question type is 22, Ryuji will change the ith book's knowledge to a new value.

Input

First line contains two integers nn and qq (nn, q \le 100000q≤100000).

The next line contains n integers represent a[i]( a[i] \le 1e9)a[i](a[i]≤1e9) .

Then in next qq line each line contains three integers aa, bb, cc, if a = 1a=1, it means question type is 11, and bb, cc represents [ ll , rr ]. if a = 2a=2 , it means question type is 22 , and bb, cc means Ryuji changes the bth book' knowledge to cc

Output

For each question, output one line with one integer represent the answer.

样例输入复制

5 3
1 2 3 4 5
1 1 3
2 5 0
1 4 5

样例输出复制

10
8

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

 

题意：两种操作 一种是更新某节点

一种是查询l-r的一个值 这个值的计算公式是：

思路：

看上去就应该是一个线段树 区间查询单点更新

但是对于查询的处理没有那么简单

可以采用前缀和的思想 因为这个公式中的系数是递减的

那么我们在线段树中维护两个值 一个是本身的值 一个是（n-i+1）倍的值

那么我们查询的时候只需要查到倍数之和 减去 本身之和的（n-r）倍就可以了

WA了一会 因为没用long long

还是要注意啊这些细节 虽然单个没有超int 但是相加就会超的

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<vector>
#include<set>
//#include<bits/stdc++.h>
#define inf 0x7f7f7f7f7f7f7f7f
using namespace std;
typedef long long LL;

const int maxn = 1e5 + 10;
LL tree[maxn << 2], treetime[maxn << 2], a[maxn];
int n, q;

void pushup(int rt)
{
	tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
	treetime[rt] = treetime[rt << 1] + treetime[rt << 1 | 1];
}

void build(int rt, int l, int r)
{
	if (l == r) {
		tree[rt] = a[l];
		treetime[rt] = a[l] * (n - l + 1);
		return;
	}
	int m = (l + r) / 2;
	build(rt << 1, l, m);
	build(rt << 1 | 1, m + 1, r);
	pushup(rt);
}

void update(int x, LL val, int l, int r, int rt)
{
	if (l == r) {
		tree[rt] = val;
		treetime[rt] = val * (n - l + 1);
		return;
	}
	int m = (l + r) / 2;
	if (x <= m) {
		update(x, val, l, m, rt << 1);
	}
	else {
		update(x, val, m + 1, r, rt << 1 | 1);
	}
	pushup(rt);
}

LL query(int L, int R, int l, int r, int rt)
{
	if (L <= l && R >= r) {
		return tree[rt];
	}
	int m = (l + r) / 2;
	LL ans = 0;
	if (L <= m) {
		ans += query(L, R, l, m, rt << 1);
	}
	if (R > m) {
		ans += query(L, R, m + 1, r, rt << 1 | 1);
	}
	pushup(rt);
	return ans;
}

LL querytime(int L, int R, int l, int r, int rt)
{
	if (L <= l && R >= r) {
		return treetime[rt];
	}
	int m = (l + r) / 2;
	LL ans = 0;
	if (L <= m) {
		ans += querytime(L, R, l, m, rt << 1);
	}
	if (R > m) {
		ans += querytime(L, R, m + 1, r, rt << 1 | 1);
	}
	return ans;
}

void init()
{
    memset(tree, 0, sizeof(tree));
    memset(treetime, 0, sizeof(treetime));
}

int main()
{
	while (scanf("%d%d", &n, &q) != EOF) {
	    if(n == 0){
            while(q--){
                int op, l, r;
                scanf("%d%d%d", &op, &l, &r);
                printf("0\n");
            }
            continue;
	    }
        init();
		for (int i = 1; i <= n; i++) {
			scanf("%lld", &a[i]);
		}
		build(1, 1, n);

		for (int i = 0; i < q; i++) {
			int op, l, r;
			scanf("%d%d%d", &op, &l, &r);
			if (op == 1) {
				LL ans = querytime(l, r, 1, n, 1) - (n - r) * query(l, r, 1, n, 1);
				printf("%lld\n", ans);
			}
			else {
				update(l, r, 1, n, 1);
			}
		}
	}
	return 0;
}