区间树 [成段更新] POJ 3468 A Simple Problem with Integers

可参考这个题

http://blog.csdn.net/gaotong2055/article/details/9300141

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

#include <iostream>
#include <stdio.h>
using namespace std;
int n,q;
long long tree[400000];
long long add[400000];

int a,b,c;
void build(int l, int r, int k){
add[k] = 0;
if(l==r){
scanf("%lld", &tree[k]);
return;
}
int m = (l+r)/2;
build( l, m, 2*k);
build( m+1, r, 2*k + 1);
tree[k] = tree[2*k] + tree[2*k+1];
}

void down(int k,int m){
if(add[k]){
add[k*2+1] += add[k];
add[k*2] += add[k];
tree[k*2] += (m-m/2) * add[k];
tree[k*2+1] += (m/2) * add[k];
add[k] = 0;
}
}

void update(int l,int r, int k){
if( a <= l && b >= r){ //找到合适的区间
add[k] += c;
tree[k] += (long long)(r - l + 1) * c;
return;
}
down(k, r-l+1); //更新子树
int m = (l+r)/2;
if(a <= m) update(l, m, 2*k);
if(b > m) update(m+1, r, 2*k+1);
tree[k] = tree[2*k] + tree[2*k+1];
}

long long query(int l, int r, int k){
if(a <= l &&  b >= r)
{
return tree[k];
}
down(k, r-l+1); //查询的时候需要更新子树
long long ans = 0;
int m = (l+r)/2;
if(a <= m)
ans += query(l , m, k*2);
if(b > m)
ans += query(m+1, r, k*2+1);
return ans;
}
int main() {
//freopen("in.txt", "r", stdin);
char cmd[5];
while(scanf("%d %d", &n, &q) != EOF){
build(1, n ,1);
while(q--){
scanf("%s", cmd);
if(cmd[0] == 'Q'){
scanf("%d %d",&a,&b);
printf("%lld\n", query(1,n,1));
}else{
scanf("%d %d %d",&a, &b, &c);
update(1, n, 1);
}
}
}
return 0;
}