A Simple Problem with Integers (线段树区间更新，区间查询 模板)

   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.

线段树的区间更新和区间查询，带lazy标记，是个模板题

#include <string.h>
#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
//区间更新，区间查询
#define ll long long
const int maxn=1e5+10;
ll sum[maxn<<2],lazy[maxn<<2];
int N,Q;
void pushdown(int rt,int l)
{
	if (lazy[rt])
	{
		lazy[rt<<1]+=lazy[rt];
		lazy[rt<<1|1]+=lazy[rt];
		sum[rt<<1]+=lazy[rt]*(l-l/2);
		sum[rt<<1|1]+=lazy[rt]*(l/2);
		lazy[rt]=0;
	}
}
void build(int l,int r,int rt)
{
	lazy[rt]=0;//
	if (l==r)
	{
		scanf("%lld",&sum[rt]);//
		//sum[rt]=a[l];
		return;
	}
	int mid=(l+r)/2;
	build(l,mid,rt*2);
	build(mid+1,r,rt*2+1);
	sum[rt]=sum[rt*2]+sum[rt*2+1];
}
void update(int L,int R,ll v,int l,int r,int rt)
{
	if (L<=l&&r<=R)//
	{
		sum[rt]+=v*(r-l+1);//!
		lazy[rt]+=v;
		return;
	}
	pushdown(rt,r-l+1);//向下更新 
	int mid=(l+r)/2;
	if (L<=mid)//!!
		update(L,R,v,l,mid,rt*2);
	if (R>mid)//!!
		update(L,R,v,mid+1,r,rt*2+1);
	sum[rt]=sum[rt*2]+sum[rt*2+1];
}
ll query(int L,int R,int l,int r,int rt)
{
	if (L<=l&&r<=R)
		return sum[rt];
	pushdown(rt,r-l+1);
	ll ans=0;
	int mid=(l+r)/2;
	if (L<=mid)
		ans+=query(L,R,l,mid,rt*2);
	if (R>mid)
		ans+=query(L,R,mid+1,r,rt*2+1);
	return ans;
}
int main()
{
	int i,j;
	int x,y;
	ll z;
	char c[2];
	while(scanf("%d%d",&N,&Q)!=EOF)
	{
		memset(sum,0,sizeof(sum));
		memset(lazy,0,sizeof(lazy));
		build(1,N,1);
		while(Q--)
		{
			scanf("%s",c);
			if (c[0]=='C')
			{
				scanf("%d%d%lld",&x,&y,&z);
				update(x,y,z,1,N,1);
			}
			else if(c[0]=='Q')
			{
				scanf("%d%d",&x,&y);
				printf("%lld
",query(x,y,1,N,1)); 
			}
		}
	}
	return 0;
}