poj 3468 A Simple Problem with Integers(线段树，区间更新）

链接：http://poj.org/problem?id=3468

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 104996 Accepted: 32769Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... ,A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁,A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... ,A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

The sums may exceed the range of 32-bit integers.

线段树，区间更新，模板，wa了一次也是因为getchar() TAT

代码：

#define _CRT_SBCURE_MO_DEPRECATE    #include<iostream>    #include<stdlib.h>    #include<stdio.h>    #include<cmath>    #include<algorithm>    #include<string>    #include<string.h>    #include<set>    #include<queue>    #include<stack>    #include<functional>     using namespace std;const int maxn = 10000000 + 10;const int INF = 0x3f3f3f3f;typedef long long ll;int a[maxn];int t, n, q;ll x, y, v, anssum;char m;struct node {ll l, r;ll addv, sum;}tree[maxn << 2];void maintain(int id){if (tree[id].l >= tree[id].r)return;tree[id].sum = tree[id << 1].sum + tree[id << 1 | 1].sum;}void pushdown(int id){if (tree[id].l >= tree[id].r)return;if (tree[id].addv) {int tmp = tree[id].addv;tree[id << 1].addv += tmp;tree[id << 1 | 1].addv += tmp;tree[id << 1].sum += (tree[id << 1].r - tree[id << 1].l + 1)*tmp;tree[id << 1 | 1].sum += (tree[id << 1 | 1].r - tree[id << 1 | 1].l + 1)*tmp;tree[id].addv = 0;}}void build(int id, ll l, ll r){tree[id].l = l;tree[id].r = r;tree[id].addv = 0;if (l == r){tree[id].sum = a[l];return;}ll mid = (l + r) >> 1;build(id << 1, l, mid);build(id << 1 | 1, mid + 1, r);maintain(id);}void updateAdd(int id, ll l, ll r, ll val){if (tree[id].l >= l && tree[id].r <= r){tree[id].addv += val;tree[id].sum += (tree[id].r - tree[id].l + 1)*val;return;}pushdown(id);ll mid = (tree[id].l + tree[id].r) >> 1;if (l <= mid)updateAdd(id << 1, l, r, val);if (mid < r)updateAdd(id << 1 | 1, l, r, val);maintain(id);}void query(int id, ll l, ll r){if (tree[id].l >= l && tree[id].r <= r) {anssum += tree[id].sum;return;}pushdown(id);ll mid = (tree[id].l + tree[id].r) >> 1;if (l <= mid)query(id << 1, l, r);if (mid < r)query(id << 1 | 1, l, r);maintain(id);}int main() {scanf("%d %d", &n, &q);memset(a, 0, sizeof(a));for (int i = 1; i <= n; i++)scanf("%d", &a[i]);build(1, 1, n);for (int j = 1; j <= q; j++) {getchar();scanf("%c", &m);if (m == 'C') {scanf("%lld %lld %lld", &x, &y, &v);updateAdd(1, x, y, v);}if (m == 'Q') {anssum = 0;scanf("%lld %lld", &x, &y);query(1, x, y);printf("%lld\n", anssum);}}return 0;}