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.

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.

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

field=source&key=POJ+Monthly--2007.11.25">POJ Monthly--2007.11.25, Yang Yi



题目意思：给定Q (1 ≤ Q ≤ 100,000)个数A1,A2 … AQ,， 以及可能多次进行的两个操作:

1)对某个区间Ai … Aj的每一个数都加n(n可变） 2) 求某个区间Ai … Aj的数的和。

就是线段树的更新与查询操作，这里要用到延迟更新。

注意sum和add都要设成int64型，由于在更新过程中。add也可能会由于累加而超出int型的范围。

#include<stdio.h>

#include<string.h>

#define M 100005

struct tree{

	int l,r;

	__int64 sum,add;

}tree[M<<2];

void pushup(int root)

{

	if(tree[root].l==tree[root].r)return;

	tree[root].sum=tree[root<<1].sum+tree[root<<1|1].sum;

	return;

}

void pushdown(int root)

{

	if(tree[root].l==tree[root].r)return;

	if(tree[root].add==0)return;

	tree[root<<1].add+=tree[root].add;

	tree[root<<1|1].add+=tree[root].add;

	tree[root<<1].sum=tree[root<<1].sum+(tree[root<<1].r-tree[root<<1].l+1)*tree[root].add;

	tree[root<<1|1].sum=tree[root<<1|1].sum+(tree[root<<1|1].r-tree[root<<1|1].l+1)*tree[root].add;

	tree[root].add=0;

	return;

}

void build(int l,int r,int root)

{

	tree[root].l=l;

	tree[root].r=r;

	tree[root].sum=0;

	tree[root].add=0;

	if(l==r){

		tree[root].sum=0;

		return;

	}

	int mid=l+r>>1;

	build(l,mid,root<<1);

	build(mid+1,r,root<<1|1);

	pushup(root);

}

void update(int l,int r,int root,__int64 z)

{

	if(tree[root].l==l&&tree[root].r==r)

	{

		tree[root].sum=tree[root].sum+(r-l+1)*z;

		tree[root].add=tree[root].add+z;    //有可能出现多次的更新操作。所以要将全部的add都累加起来。

		return;

	}

	pushdown(root);

	int mid=tree[root].l+tree[root].r>>1;

	if(r<=mid)update(l,r,root<<1,z);

	else if(l>mid)update(l,r,root<<1|1,z);

	else {

		update(l,mid,root<<1,z);

		update(mid+1,r,root<<1|1,z);

	}

	pushup(root);

	return;

}

__int64  Query(int l,int r,int root)

{

	if(l==tree[root].l&&r==tree[root].r){

		return tree[root].sum;

	}

	pushdown(root);

	int mid=tree[root].l+tree[root].r>>1;

	if(r<=mid)return Query(l,r,root<<1);

	else if(l>mid)return Query(l,r,root<<1|1);

	else {

		return Query(l,mid,root<<1)+Query(mid+1,r,root<<1|1);

	}

}

int main()

{

	int N,Q,i,j,k,a,b;

	__int64 c,d;

	char s[20];

	while(scanf("%d%d",&N,&Q)!=EOF)

	{

		build(1,N,1);

	 for(i=1;i<=N;i++)

	 {

 		scanf("%I64d",&d);

 		update(i,i,1,d);

 	}

 	while(Q--)

	{

	scanf("%s%d%d",s,&a,&b);

	if(s[0]=='Q'){

		printf("%I64d\n",Query(a,b,1));

	}

	if(s[0]=='C'){

		scanf("%I64d",&c);

		update(a,b,1,c);

	}

	}

	}

	return 0;

}