88 lines
2.5 KiB
C++
88 lines
2.5 KiB
C++
/*
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题目：
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将某区间每一个数加上k
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求出某区间每一个数的和
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题目总结：
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区间修改，区间查询
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*/
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#include <iostream>
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using namespace std;
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#define LL long long
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#define N 100010
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//快读
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LL read() {
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LL 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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int n, q, m;
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LL tr[N << 2], tag[N << 2];
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void build() {
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for (m = 1; m <= n;) m <<= 1;
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for (int i = 1; i <= n; i++) tr[m + i] = read();
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for (int i = m - 1; i >= 1; i--) tr[i] = tr[i << 1] + tr[i << 1 | 1];
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}
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void add(int l, int r, int d) {
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int lc = 0, rc = 0, c = 1;
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// 开区间，从左右叶子端点出发，如果l和r不是兄弟节点，就一路向上
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// lc 表示当前左端点走到的子树有多少个元素在修改区间内
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// rc 表示当前右端点走到的子树有多少个元素在修改区间内
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// c 当前层叶子节点个数
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for (l = m + l - 1, r = m + r + 1; l ^ r ^ 1; l >>= 1, r >>= 1, c <<= 1) {
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tr[l] += d * lc, tr[r] += d * rc; //自底向上，更新统计信息tr和
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if (~l & 1) tag[l ^ 1] += d, tr[l ^ 1] += d * c, lc += c; // l % 2 == 0 左端点 是左儿子 右子树完整在区间内
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if (r & 1) tag[r ^ 1] += d, tr[r ^ 1] += d * c, rc += c; // r % 2 == 1 右端点 是右儿子 左子树完整在区间内
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}
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//是兄弟节点时，需要一路向上到根进行推送
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for (; l || r; l >>= 1, r >>= 1) tr[l] += d * lc, tr[r] += d * rc;
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}
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//区间查询
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LL query(int l, int r) {
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LL lc = 0, rc = 0, c = 1;
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LL res = 0;
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for (l = m + l - 1, r = m + r + 1; l ^ r ^ 1; l >>= 1, r >>= 1, c <<= 1) {
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if (tag[l]) res += tag[l] * lc;
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if (tag[r]) res += tag[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 || r; l >>= 1, r >>= 1) res += tag[l] * lc, res += tag[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("P3372.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 (int i = 1; i <= q; i++) {
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int op = read();
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int l = read(), r = read();
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if (op == 1) {
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int c = read();
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add(l, r, c);
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} else
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printf("%lld\n", query(l, r));
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}
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return 0;
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} |