poj 3468 A Simple Problem with Integers

时间：2014-04-28 10:23:41 收藏：0 阅读：282

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 55626		Accepted: 16755
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

Hint

The sums may exceed the range of 32-bit integers.

线段树+延迟标记

#include"stdio.h"
#include"string.h"
#define N 100005
int num[N];
struct node
{
	int l,r;
	__int64 s,a;
}f[N*3];
void creat(int t,int l,int r)
{
	f[t].l=l;
	f[t].r=r;
	f[t].a=0;
	if(l==r)
	{
		f[t].s=num[r];
		return ;
	}
	int temp=t*2,mid=(l+r)/2;
	creat(temp,l,mid);
	creat(temp+1,mid+1,r);
	f[t].s=f[temp].s+f[temp+1].s;
	return ;
}
void update(int t,int l,int r,__int64 a)
{
	if(f[t].l==l&&f[t].r==r)
	{
		f[t].a+=a;           //每次都是s先增加，a记录下一层的增量
		f[t].s+=(f[t].r-f[t].l+1)*a;
		return ;       //不再向下一层更新
	}
	int temp=t*2,mid=(f[t].l+f[t].r)/2;
	if(f[t].a)
	{
		f[temp].a+=f[t].a;
		f[temp].s+=(f[temp].r-f[temp].l+1)*f[t].a;
		f[temp+1].a+=f[t].a;
		f[temp+1].s+=(f[temp+1].r-f[temp+1].l+1)*f[t].a;
		f[t].a=0;            //完成向下一层的加和，进行归零
	}
	if(r<=mid)
		update(temp,l,r,a);
	else if(l>mid)
		update(temp+1,l,r,a);
	else
	{
		update(temp,l,mid,a);
		update(temp+1,mid+1,r,a);
	}
	f[t].s=f[temp].s+f[temp+1].s;
	return ;
}
__int64 find(int t,int l,int r)
{
	if(f[t].l==l&&f[t].r==r)
	{
		return f[t].s;
	}
	int temp=t*2,mid=(f[t].l+f[t].r)/2;
	if(f[t].a)              //下一层的更新还没完成
	{
		f[temp].a+=f[t].a;
		f[temp].s+=(f[temp].r-f[temp].l+1)*f[t].a;
		f[temp+1].a+=f[t].a;
		f[temp+1].s+=(f[temp+1].r-f[temp+1].l+1)*f[t].a;
		f[t].a=0;            
	}
	if(r<=mid)
		return find(temp,l,r);
	else if(l>mid)
		return find(temp+1,l,r);
	else
		return find(temp,l,mid)+find(temp+1,mid+1,r);
}
int main()
{
	int i,n,q;
	int a,b,c;
	char ch;
	while(scanf("%d%d",&n,&q)!=-1)
	{
		for(i=1;i<=n;i++)
			scanf("%d",&num[i]);
		creat(1,1,n);
		while(q--)
		{
			getchar();
			scanf("%c",&ch);
			if(ch==‘C‘)
			{
				scanf("%d%d%d",&a,&b,&c);
				update(1,a,b,c);
			}
			else
			{
				scanf("%d%d",&a,&b);
				printf("%I64d\n",find(1,a,b));
			}
		}
	}
	return 0;
}

poj 3468 A Simple Problem with Integers,码迷,mamicode.com