poj 3468 A Simple Problem with Integers （线段树成段更新）

                                                                  A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 77486		Accepted: 23862
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

题目链接：http://poj.org/problem?id=3468

题目大意：C情况时a-b区间内每一个数字加入c。Q情况时查询区间内值的和。

解题思路：模板线段树成段更新问题。

代码例如以下：

#include <cstdio>
#include <cstring>
#define inf 1e12
#define ll long long 
const ll maxn=1000010;
ll n,m,a,b,c;
ll sum[maxn],nd[maxn];
char s[2];
void build(ll l,ll r,ll root)
{
	ll mid=(l+r)/2;
	if(l==r)
	{
		scanf("%lld",&sum[root]);
		return;
	}
	build(l,mid,root*2);
	build(mid+1,r,root*2+1);
	sum[root]=sum[root*2]+sum[root*2+1];
}
void pushdown(ll l,ll root)
{
	if(nd[root]!=0)
	{
		nd[root*2]+=nd[root];
		nd[root*2+1]+=nd[root];
		sum[root*2]+=((l-l/2)*nd[root]);
		sum[root*2+1]+=((l/2)*nd[root]);
		nd[root]=0;
	}
}
ll query(ll l,ll r,ll root)
{
	if(l>=a&&r<=b)
		return sum[root];
	pushdown(r-l+1,root);
	ll mid=(l+r)/2,ans=0;
	if(a<=mid)
		ans+=query(l,mid,root*2);
	if(b>mid)
		ans+=query(mid+1,r,root*2+1);
	return ans;			
}
void update(ll l,ll r,ll root)
{
	if(l>b||r<a)
		return;
	ll len=r-l+1;
	if(l>=a&&r<=b)
	{
		nd[root]+=c;
		sum[root]+=(len*c);
		return ;
	}
	pushdown(len,root);
	int mid=(l+r)/2;
	if(a<=mid)
	update(l,mid,root*2);
	if(b>mid)
	update(mid+1,r,root*2+1);
	sum[root]=sum[root*2]+sum[root*2+1];
}
int main(void)
{
	memset(nd,0,sizeof(nd));
	scanf("%lld%lld",&n,&m);
	build(1,n,1);
	for(ll i=0;i<m;i++)
	{
		scanf("%s",s);
		if(s[0]=='Q')
		{
			scanf("%lld%lld",&a,&b);
			printf("%lld\n",query(1,n,1));
		}
		else
		{
			scanf("%lld%lld%lld",&a,&b,&c);
			update(1,n,1);
		}
	}
}