poj 3468

A Simple Problem with Integers

Time Limit: 5000MS

 

Memory Limit: 131072K

Total Submissions: 90736

 

Accepted: 28268

Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. You need to dealwith two kinds of operations. One type of operation is to add some given numberto each number in a given interval. The other is to ask for the sum of numbersin a given interval.

Input

The first linecontains 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 queryingthe sum of A_a, A_a+1, ... , A_b.

Output

You need to answerall Q commands in order. One answer in a line.

SampleInput

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

SampleOutput

4

55

9

15

Hint

The sumsmay exceed(超过) the range of 32-bit integers(整数).

Source

POJMonthly--2007.11.25, Yang Yi

 注意：数组要用long long

 线段树解法

Problem: 3468

User: ksq2013

Memory: 4776K

Time: 2532MS

Language: G++

Result: Accepted

#include<cstdio>#include<cstdlib>#include<iostream>using namespace std;int n,q;long long col[400001],sum[400001];void pushup(int k){sum[k]=sum[k<<1]+sum[k<<1|1];}void pushdown(int k,int m){if(col[k]){col[k<<1]+=col[k];col[k<<1|1]+=col[k];sum[k<<1]+=col[k]*(m-(m>>1));sum[k<<1|1]+=col[k]*(m>>1);col[k]=0;}}void build(int s,int t,int k){if(!(s^t)){scanf("%lld",&sum[k]);return;}int m=(s+t)>>1;build(s,m,k<<1);build(m+1,t,k<<1|1);pushup(k);}void update(int s,int t,int k,int l,int r,int c){if(l<=s&&t<=r){col[k]+=c;sum[k]+=c*(t-s+1);return;}pushdown(k,t-s+1);int m=(s+t)>>1;if(l<=m)update(s,m,k<<1,l,r,c);if(m<r)update(m+1,t,k<<1|1,l,r,c);pushup(k);}long long query(int s,int t,int k,int l,int r){if(l<=s&&t<=r)return sum[k];pushdown(k,t-s+1);int m=(s+t)>>1;long long res=0;if(l<=m)res+=query(s,m,k<<1,l,r);if(m<r)res+=query(m+1,t,k<<1|1,l,r);return res;}int main(){scanf("%d%d",&n,&q);build(1,n,1);for(int i=1;i<=q;i++){char ak[3];scanf("%s",ak);if(ak[0]=='C'){int a,b,c;scanf("%d%d%d",&a,&b,&c);update(1,n,1,a,b,c);}else{int a,b;scanf("%d%d",&a,&b);printf("%lld\n",query(1,n,1,a,b));}}return 0;}