一、使用树状数组实现
区间修改和区间查询
我们还是利用差分，可以得：
\(\displaystyle \sum_{i=1}^na_i = \sum_{i=1}^n \sum_{j=1}^i b_j\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ = n \times b_1 + (n-1) \times b_2 + ... + b_n\)
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ = n \times \sum_{i=1}^n b_i - \sum_{i=1}^nb_i \times (i-1)\)
方法就显而易见了：维护两个树状数组，一个维护 \(\displaystyle \sum_{i=1}^nb_i\),一个维护\(\displaystyle \sum_{i=1}^nb_i \times (i-1)\)
实现代码
#include <bits/stdc++.h>
using namespace std;
#define lowbit(x) (x & -x)
typedef long long LL;
#define N 1000010
LL n, q, op, l, r, x;
LL a[N];
LL t1[N], t2[N];
LL add(LL x, LL y) {
for (LL i = x; i <= n; i += lowbit(i)) {
t1[i] += y;
t2[i] += x * y;
}
}
LL ask(LL x) {
LL ans = 0;
for (LL i = x; i; i -= lowbit(i)) ans += (x + 1) * t1[i] - t2[i];
return ans;
}
int main() {
scanf("%lld%lld", &n, &q);
for (int i = 1; i <= n; i++) {
scanf("%lld", &a[i]);
add(i, a[i] - a[i - 1]);
}
for (int i = 1; i <= q; i++) {
scanf("%d%lld%lld", &op, &l, &r);
if (op == 1) {
scanf("%lld", &x);
add(l, x);
add(r + 1, -x);
} else
printf("%lld\n", ask(r) - ask(l - 1));
}
return 0;
}