poj3468 A Simple Problem with Integers 线段树入门题复习

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 93133		Accepted: 28985
Case Time Limit: 2000MS

Description

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.

思路：跟上一题差不多，但是注意查询的时候要判断有木有懒惰标记，继续更新。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn=100005;
LL sum[4*maxn];
LL col[4*maxn];
void buildtree(int l,int r,int pos)
{
col[pos]=sum[pos]=0;
if(l==r){scanf("%lld",&sum[pos]);return;}
int mid=(l+r)>>1;
buildtree(l,mid,pos<<1);
buildtree(mid+1,r,pos<<1|1);
sum[pos]=sum[pos<<1]+sum[pos<<1|1];
}
void update(int l,int r,int pos,int resl,int resr,int value)
{
//不用怕前面积攒了很多，因为遍历到父亲的时候你已经更新了
if(l==r){sum[pos]+=value; return;}
if(resl<=l&&r<=resr)
{
sum[pos]+=value*(r-l+1);
col[pos]+=value;//儿子们需要增加的值
return;
}
int mid=(l+r)>>1;
if(col[pos])
{
col[pos<<1]+=col[pos];
col[pos<<1|1]+=col[pos];
sum[pos<<1]+=col[pos]*(mid-l+1);
sum[pos<<1|1]+=col[pos]*(r-mid);
col[pos]=0;
}
if(resl<=mid)update(l,mid,pos<<1,resl,resr,value);
if(resr>mid)update(mid+1,r,pos<<1|1,resl,resr,value);
sum[pos]=sum[pos<<1]+sum[pos<<1|1];
}
LL query(int l,int r,int pos,int ql,int qr)
{
LL ans=0;
if(ql<=l&&r<=qr){return sum[pos];}
int mid=(l+r)>>1;
if(col[pos])
{
col[pos<<1]+=col[pos];
col[pos<<1|1]+=col[pos];
sum[pos<<1]+=col[pos]*(mid-l+1);
sum[pos<<1|1]+=col[pos]*(r-mid);
col[pos]=0;//儿子已经更新完毕
}
if(ql<=mid)ans+=query(l,mid,pos<<1,ql,qr);
if(qr>mid)ans+=query(mid+1,r,pos<<1|1,ql,qr);
return ans;
}
int main()
{
int n,m,a,b,c;
char q[10];
while(scanf("%d%d",&n,&m)==2)
{
buildtree(1,n,1);
while(m--)
{
scanf("%s%d%d",&q,&a,&b);
if(q[0]=='Q')
{
printf("%lld\n",query(1,n,1,a,b));
}
else if(q[0]=='C')
{
scanf("%d",&c);
update(1,n,1,a,b,c);
}
}
}
}