Solution

用树状数组，代码量小，空间占用小
巧用增量数组，修改时在 (l) 处 $ + val$ ， (r + 1) 处 $ - val$，在 (x) 处的值就是 (sum_{i = 1}^{x}c[i])
这就是区间更新，单点求值的树状数组

那么怎么区间更新区间查询呢？
设增量数组为 (b[i])
显然，查询 (1 - x) 的答案为： (sum_{i = 1}^{x}sum_{j = 1}^{i}b[i])
这样还不明朗，无法得到一个比较通项的前缀和，所以进行如下变换：

[sum_{i = 1}^{x}sum_{j = 1}^{i}b[i]$$$$=sum_{i = 1}^{x}(x - i + 1) * b[i]$$$$=(x + 1)sum_{i = 1}^{x}b[i] - sum_{i = 1}^{x}i * b[i] ]

这样一来，我们通过两个树状数组分别维护 (sum_{i = 1}^{x}b[i]) 和 (sum_{i = 1}^{x}i * b[i])，即可快速计算得答案

具体的，第一个树状数组我们在 (l) 处 $ + val$ ， (r + 1) 处 (- val)，第二个树状数组在 (l) 处 $ + l * val$， (r + 1) 处 (- (r + 1) * val)
(1-x)前缀和即为 ((x + 1) * get\_sum(x, 0) - get\_sum(x, 1))

Code

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
#include<climits>
#define LL long long
using namespace std;
LL RD(){
    LL out = 0,flag = 1;char c = getchar();
    while(c < '0' || c >'9'){if(c == '-')flag = -1;c = getchar();}
    while(c >= '0' && c <= '9'){out = out * 10 + c - '0';c = getchar();}
    return flag * out;
    }
const LL maxn = 400019;
LL num, na;
#define lowbit(i) ((i) & (-i))
LL v[maxn], sum[maxn];
LL c[maxn][2];
void update(LL x, LL val, LL o){
	for(LL i = x;i <= num;i += lowbit(i))c[i][o] += val;
	}
LL get_sum(LL x, LL o){
	LL ans = 0;
	for(LL i = x;i > 0;i -= lowbit(i))ans += c[i][o];
	return ans;
	}
int main(){
	num = RD();
	for(LL i = 1;i <= num;i++)v[i] = RD(), sum[i] = sum[i - 1] + v[i];
	na = RD();
	for(LL i = 1;i <= na;i++){
		LL cmd = RD(), l = RD(), r = RD();
		if(cmd == 1){
			LL val = RD();
			update(l, val, 0), update(r + 1, -val, 0);
			update(l, l * val, 1), update(r + 1, (r + 1) * -val, 1);
			}
		else{
			printf("%lld
",sum[r] - sum[l - 1] + (r + 1) * get_sum(r, 0) - get_sum(r, 1) - l * get_sum(l - 1, 0) + get_sum(l - 1, 1));
			}
		}
	}