poj 3468 A Simple Problem with Integers（线段树、延迟更新）

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 74705		Accepted: 22988
Case 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 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.

Source

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
",Query(a,b,1));
	}
	if(s[0]=='C'){
		scanf("%I64d",&c);
	
		update(a,b,1,c);
	
	}
	}
	}
	return 0;
}