POJ A Simple Problem with Integers

http://poj.org/problem?id=3468

 

 

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 34676		Accepted: 9917
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, ... , 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

 

 

#include<stdio.h>
typedef __int64 LL;
struct SegmentTree{
    int left,right;
    LL  sum,add;
}f[600000];
LL value[100100];
LL build(int left,int right,int k){
    f[k].left=left,    f[k].right=right;
    f[k].add = 0;
    if(left==right){
        return f[k].sum = value[left];
    }    
    int mid= (left+right)>>1;
    return f[k].sum=build(left,mid,k<<1)+build(mid+1,right,k<<1|1);
}
void update(int left,int right,LL add,int k){
    if(left==f[k].left&&right==f[k].right){
        f[k].add+=add;
        f[k].sum+=(f[k].right-f[k].left+1)*add;
        return ;
    }
    if(f[k].add){
        f[k<<1].add+=f[k].add;
        f[k<<1|1].add+=f[k].add;
        f[k<<1].sum+=(f[k<<1].right-f[k<<1].left+1)*f[k].add;
        f[k<<1|1].sum+=(f[k<<1|1].right-f[k<<1|1].left+1)*f[k].add;
        f[k].add=0;
    }
    int mid = (f[k].left+f[k].right)>>1;
    if(right<=mid)
        update(left,right,add,k<<1);
    else    if(left>mid)
                update(left,right,add,k<<1|1);
            else{
                update(left,mid,add,k<<1);
                update(mid+1,right,add,k<<1|1);
            }
    f[k].sum=f[k<<1].sum+f[k<<1|1].sum;
}
LL query(int left,int right,int k){
    if(f[k].left==left&&f[k].right==right)
        return f[k].sum;
    if(f[k].add){
        f[k<<1].add+=f[k].add;
        f[k<<1|1].add+=f[k].add;
        f[k<<1].sum+=(f[k<<1].right-f[k<<1].left+1)*f[k].add;
        f[k<<1|1].sum+=(f[k<<1|1].right-f[k<<1|1].left+1)*f[k].add;
        f[k].add=0;
    }
    int mid = (f[k].left+f[k].right)>>1;
    if(right<=mid)
        return query(left,right,k<<1);
    else    if(left>mid)
                return query(left,right,k<<1|1);
            else
                return query(left,mid,k<<1)+query(mid+1,right,k<<1|1);
}
int main(){
    int n,q;
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;i++)
        scanf("%I64d",&value[i]);
    build(1,n,1);
    while(q--){
        char s[2];
        scanf("%s",s);
        if(s[0]=='Q'){
            int left,right;
            scanf("%d%d",&left,&right);
            printf("%I64d\n",query(left,right,1));
        }
        else{
            int left,right;
            LL add;
            scanf("%d%d%I64d",&left,&right,&add);
            update(left,right,add,1);
        }
    }
    return 0;
}