P3368【模板】树状数组 2 - 差分

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建立两个差分数组，套公式就好了
c[i]表示i元素的“增量”，下面的式子左边是序列从1 ~ x的前缀和整体增加的值
$\sum_{i=1}^x\sum_{j=1}^ic[j] = (x+1)\sum_{i=1}^xc[i] - \sum_{i=1}^xi*c[i]$

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
#define debug(x) cerr << #x << "=" << x << endl;
const int MAXN = 500000 + 10;
int n,m,tr[MAXN],c1[MAXN],c2[MAXN],sum[MAXN];
void update(int k, int p, int c[]) {
	while(p <= n) {
		c[p] += k;
		p += p&(-p);
	}
}
int getsum(int p, int c[]) {
	int sum = 0;
	while(p) {
		sum += c[p];
		p -= p&(-p);
	}
	return sum;
}
int main() {
	scanf("%d%d", &n, &m);
	for(int i=1; i<=n; i++) {
		int ai = 0;
		scanf("%d", &ai);
		sum[i] = sum[i-1] + ai;
	}
	for(int i=1; i<=m; i++) {
		int cmd, x, y, k;
		scanf("%d", &cmd);
		if(cmd == 1) {
			scanf("%d%d%d", &x, &y, &k);
			update(k, x, c1), update(-k, y+1, c1);
			update(k*x, x, c2), update(-k * (y+1), y+1, c2);
		} else {
			scanf("%d", &x);
			int ls = sum[x-1] + x * getsum(x-1, c1) - getsum(x-1, c2);
			int rs = sum[x] + (x+1) * getsum(x, c1) - getsum(x, c2);
			printf("%d\n", rs - ls);
		}
	}
	return 0;
}