poj 3468 A Simple Problem with Integers(线段树区间更新)

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 122316		Accepted: 37938
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.
Source

POJ Monthly--2007.11.25, Yang Yi

题意：有n个数q次操作，每次操作为Q时，是查询l到r的区间的和。操作为C时，是从l到r的每个数都加上一个值。

思路：线段树区间更新。注意懒惰标记也要开long long，不然会WA。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=100000;
long long tr[4*maxn];
long long add[4*maxn];
void pushup(int i)
{
tr[i]=tr[i<<1]+tr[i<<1|1];
}
void pushdown(int i,int len)
{
if(add[i])
{
add[i<<1]+=add[i];
add[i<<1|1]+=add[i];
tr[i<<1]+=add[i]*(len-len/2);
tr[i<<1|1]+=add[i]*(len/2);
add[i]=0;
}
}
void build(int i,int l,int r)
{
if(l==r)
{
scanf("%lld",&tr[i]);
return;
}
int mid=(l+r)/2;
build(i*2,l,mid);
build(i*2+1,mid+1,r);
pushup(i);
}
void update(int i,int l,int r,int x,int y,int c)
{
if(x<=l&&r<=y)
{
add[i]+=c;
tr[i]=tr[i]+(r-l+1)*c;
return;
}
pushdown(i,r-l+1);
int mid=(l+r)/2;
if(x<=mid) update(2*i,l,mid,x,y,c);
if(y>mid) update(2*i+1,mid+1,r,x,y,c);
pushup(i);
}
long long query(int i,int l,int r,int x,int y)
{
long long sum=0;
if(x<=l&&r<=y)
{
sum+=tr[i];
return sum;
}
pushdown(i,r-
bfa2
l+1);
int mid=(l+r)/2;
if(x<=mid) sum+=query(2*i,l,mid,x,y);
if(y>mid) sum+=query(2*i+1,mid+1,r,x,y);
pushup(i);
return sum;
}
int main()
{
int n,q,x,y,c;
char str[10];
while(~scanf("%d%d",&n,&q))
{
build(1,1,n);
memset(add,0,sizeof(add));
for(int i=1;i<=q;i++)
{
scanf("%s",str);
if(str[0]=='Q')
{
scanf("%d%d",&x,&y);
printf("%lld\n",query(1,1,n,x,y));
}
else
{
scanf("%d%d%d",&x,&y,&c);
update(1,1,n,x,y,c);
}
}
}
return 0;
}