POJ3468 线段树（区间更新，区间求和，延迟标记）

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 97196		Accepted: 30348
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.
题意：
一串数，有两种操作，在区间a,b内每一个数加c,求区间a,b数的和。
线段树模板题。
代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,q;
long long sum[400005];//数组多乘4倍。
long long add[400005];
void pushup(int rt)
{
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void pushdown(int rt,int lne)
{
if(add[rt])
{
add[rt<<1]+=add[rt];
add[rt<<1|1]+=add[rt];
sum[rt<<1]+=add[rt]*(lne-(lne>>1));
sum[rt<<1|1]+=add[rt]*(lne>>1);
add[rt]=0;
}
}
void build(int l,int r,int rt)
{
add[rt]=0;
if(l==r)
{
scanf("%lld",&sum[rt]);
return;
}
int mid=(l+r)>>1;
build(l,mid,rt<<1);
build(mid+1,r,rt<<1|1);
pushup(rt);
}
void update(int L,int R,int c,int l,int r,int rt)
{
if(L<=l&&R>=r)
{
add[rt]+=c;
sum[rt]+=(long long)c*(r-l+1);
return;
}
pushdown(rt,r-l+1);
int mid=(l+r)>>1;
if(L<=mid)
update(L,R,c,l,mid,rt<<1);
if(R>mid)
update(L,R,c,mid+1,r,rt<<1|1);
pushup(rt);
}
long long query(int L,int R,int l,int r,int rt)
{
if(L<=l&&R>=r)
{
return sum[rt];
}
pushdown(rt,r-l+1);
int mid=(l+r)>>1;
long long ans=0;
if(L<=mid)
ans+=query(L,R,l,mid,rt<<1);
if(R>mid)
ans+=query(L,R,mid+1,r,rt<<1|1);
return ans;
}
int main()
{
int a,b,c;
scanf("%d%d",&n,&q);
build(1,n,1);
while(q--){
char ch[3];
scanf("%s",ch);  //不能用scanf("%c".....),字符后有空格，可以scanf一个字符数组/cin。
if(ch[0]=='Q')
{
scanf("%d%d",&a,&b);
printf("%lld\n",query(a,b,1,n,1));
}
else if(ch[0]=='C')
{
scanf("%d%d%d",&a,&b,&c);
update(a,b,c,1,n,1);
}
}
return 0;
}