POJ a simple problem of Integers

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.
 

对区间处理优化，单点处理太费时

 

 

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<cmath>
#include<stdio.h>
using namespace std;

typedef long long ll;

ll n,q;
ll a[100005],c1[100005],c2[100005];
ll x,y,z,sum1,sum2;

ll lowbit(ll i)
{
    return i&(-i);
}

void add(ll *c,ll i,ll data)
{
    while(i<=n)
    {
        c[i]+=data;
        i+=lowbit(i);
    }
}

long long getsum(ll *c,ll i)
{
    long long ans=0;
    while(i>0)
    {
        ans+=c[i];
        i-=lowbit(i);
    }
    return ans;
}

int main()
{
    int i,j;
    scanf("%lld%lld",&n,&q);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        add(c1,i,a[i]-a[i-1]);
        add(c2,i,(i-1)*(a[i]-a[i-1]));
    }
    while(q--)
    {
        char ch;
        cin>>ch;
        if(ch=='Q')
        {
            scanf("%lld%lld",&x,&y);
            sum1=(x-1)*getsum(c1,x-1)-getsum(c2,x-1);
            sum2=y*getsum(c1,y)-getsum(c2,y);
            printf("%lld\n",sum2-sum1);
        }
        if(ch=='C')
        {
            scanf("%lld%lld%lld",&x,&y,&z);
            add(c1,x,z);
            add(c1,y+1,-z);
            add(c2,x,z*(x-1));
            add(c2,y+1,-z*y);
        }
    }
    return 0;
}