poj 3468 (区间修改区间查询)

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions:147133		Accepted: 45718
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, ... , 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 querying the sum of A_a, A_a₊₁, ... , A_b.

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

Hint

The sums may exceed the range of 32-bit integers.

#include<iostream>

#include<cstdio>

#include<cmath>

#include<algorithm>

#include<cstring>

using namespace std;

int n,m;

const int N=1e5+;

long long s1[N],s2[N];

int lowbit(int x)

{

    return x&(-x);

}

void updata(int p,long long x)

{

    for(int i=p;i<=n;i+=lowbit(i)){

        s1[i]+=x;

        s2[i]+=x*p;

    }

}

long long sum(int p)

{

    long long ans=;

    for(int i=p;i>;i-=lowbit(i)){

        ans+=s1[i]*(p+)-s2[i];

    }

    return ans;

}

int main()

{

    while(scanf("%d %d",&n,&m)==){

        memset(s1,,sizeof(s1));

        memset(s2,,sizeof(s2));

        for(int i=;i<=n;i++){

            long long x;

            scanf("%lld",&x);

            updata(i,x);

            updata(i+,-x);

        }

        while(m--){

            char s[];

            scanf("%s",s);

            if(s[]=='C'){

                int a,b;

                long long c;

                scanf("%d %d %lld",&a,&b,&c);

                updata(a,c);

                updata(b+,-c);

            }

            else{

                int a,b;

                scanf("%d %d",&a,&b);

                printf("%lld\n",sum(b)-sum(a-));

            }

        }

    }

    return ;

}

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poj 3468 (区间修改区间查询)

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