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