|
|
// Luogu.org P1438
|
|
|
// https://www.cnblogs.com/cytus/p/9590071.html
|
|
|
#include <cstdio>
|
|
|
#include <cstring>
|
|
|
#include <algorithm>
|
|
|
|
|
|
using namespace std;
|
|
|
#define LL long long
|
|
|
const int N = 1e5 + 7;
|
|
|
int n, q, m;
|
|
|
LL a[N], tr[N << 2], add[N << 2];
|
|
|
|
|
|
//快读
|
|
|
int read() {
|
|
|
int x = 0, f = 1;
|
|
|
char ch = getchar();
|
|
|
while (ch < '0' || ch > '9') {
|
|
|
if (ch == '-') f = -1;
|
|
|
ch = getchar();
|
|
|
}
|
|
|
while (ch >= '0' && ch <= '9') {
|
|
|
x = (x << 3) + (x << 1) + (ch ^ 48);
|
|
|
ch = getchar();
|
|
|
}
|
|
|
return x * f;
|
|
|
}
|
|
|
|
|
|
void modify(int x, int d) {
|
|
|
for (x += m; x; x >>= 1) tr[x] += d; //单点修改，一路向上更新
|
|
|
}
|
|
|
|
|
|
void modify(int l, int r, int d) { //区间修改
|
|
|
LL lc = 0, rc = 0, c = 1;
|
|
|
for (l = l + m - 1, r = r + m + 1; l ^ r ^ 1; l >>= 1, r >>= 1, c <<= 1) {
|
|
|
tr[l] += d * lc, tr[r] += d * rc;
|
|
|
|
|
|
if (~l & 1) add[l ^ 1] += d, tr[l ^ 1] += d * c, lc += c;
|
|
|
if (r & 1) add[r ^ 1] += d, tr[r ^ 1] += d * c, rc += c;
|
|
|
}
|
|
|
//一路向上走,修改sum值，但却没有修改add
|
|
|
for (; l; l >>= 1, r >>= 1) tr[l] += d * lc, tr[r] += d * rc;
|
|
|
}
|
|
|
|
|
|
LL query(int l, int r) {
|
|
|
LL res = 0, lc = 0, rc = 0, c = 1;
|
|
|
for (l = l + m - 1, r = r + m + 1; l ^ r ^ 1; l >>= 1, r >>= 1, c <<= 1) {
|
|
|
if (add[l]) res += add[l] * lc;
|
|
|
if (add[r]) res += add[r] * rc;
|
|
|
if (~l & 1) res += tr[l ^ 1], lc += c;
|
|
|
if (r & 1) res += tr[r ^ 1], rc += c;
|
|
|
}
|
|
|
for (; l; l >>= 1, r >>= 1) res += add[l] * lc, res += add[r] * rc;
|
|
|
return res;
|
|
|
}
|
|
|
int main() {
|
|
|
//文件输入输出
|
|
|
#ifndef ONLINE_JUDGE
|
|
|
freopen("P1438.in", "r", stdin);
|
|
|
#endif
|
|
|
n = read(), q = read();
|
|
|
|
|
|
// build
|
|
|
for (m = 1; m <= n + 1;) m <<= 1;
|
|
|
for (int i = 1; i <= n; i++) {
|
|
|
a[i] = read();
|
|
|
tr[m + i] = a[i] - a[i - 1]; //记录差分值
|
|
|
}
|
|
|
|
|
|
// debug
|
|
|
// printf("m=%d\n", m);
|
|
|
// for (int i = 0; i <= 6; i++) printf("%lld ", tr[m + i]);
|
|
|
// puts("");
|
|
|
|
|
|
//所有父节点，从下向上，将值更新为左右子树的和,每个值都是自已所辖区间的差分和
|
|
|
for (int i = m - 1; i >= 1; i--) {
|
|
|
tr[i] = tr[i << 1] + tr[i << 1 | 1];
|
|
|
// printf("tr[%d]=%d\n", i, tr[i]);
|
|
|
}
|
|
|
|
|
|
/*
|
|
|
1 2 3 4 5
|
|
|
|
|
|
1 2 3 4 5
|
|
|
1 3 5
|
|
|
----------
|
|
|
1 3 6 9 5
|
|
|
|
|
|
query(3)=6
|
|
|
|
|
|
tr
|
|
|
1 1 1 1 1
|
|
|
|
|
|
*/
|
|
|
|
|
|
for (int i = 1; i <= q; i++) {
|
|
|
int op = read();
|
|
|
if (op == 1) {
|
|
|
int l = read(), r = read();
|
|
|
LL k = read(), d = read(); //首项:k,公差:d
|
|
|
// 1 2 4 1 2
|
|
|
modify(l, k); //改2
|
|
|
modify(r + 1, -(k + (r - l) * d)); //改4+1
|
|
|
modify(l + 1, r, d); //改3-4
|
|
|
} else {
|
|
|
LL x = read(); //查询单点
|
|
|
printf("%lld\n", query(1, x)); //以前缀和进行描述
|
|
|
}
|
|
|
}
|
|
|
return 0;
|
|
|
} |
