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: 131072K
Total Submissions: 104996		Accepted: 32769
Case Time Limit: 2000MS

Description

You have N integers, A1, A2, ... ,
AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.

Each of the next Q lines represents an operation.

"C a b c" means adding c to each of Aa,
Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.

"Q a b" means querying the sum of Aa, Aa+1, ... ,
Ab.

Output

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

Sample Input

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

Sample Output

4
55
9
15

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;
}