POJ 3468 A Simple Problem with Integers (树状数组) (区间修改+区间查询)

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 114049		Accepted: 35396
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.

区间值 修改,  区间查询;

首先 会超出int  范围 所以用 long long ;

 区间修改 与单点修改不同  

思路：两个数组C1[x]表示该点元素与左边的差值，C2[x]表示的是x*C[x]

 因此:  求区间的和 为 : 

1.  sum(sum(C[j],j<=i)i<=x)  
2.  =(x+1)*sum(C[i],i<=x)-sum(i*C[i],i<=x); 

区间修改[a,b]:   

add(c1,a,num);

     add(c1,b+1,-num);

     add(c2,a,a*num);

     add(c2,b+1,-num*(b+1));

区间求和[a,b]:

ll Get_sum(ll x)

{

    return Sum(x,c1)*x-Sum(x,c2);

}

ll Get(ll a,ll b)

{

    return Get_sum(b+1)-Get_sum(a);// 闭区间从 b+1 ---a;

}

AC代码:

#include <stdio.h>
#include <queue>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
typedef long long ll;
using namespace std;
const int N =1000110;
ll c1
,c2
;
ll air
;
ll n;
void add(ll tree[],ll k,ll num)
{
while(k<=n)
{
tree[k]+=num;
k+=k&(-k);
}
}
ll Sum(ll x,ll tree[])
{
ll sum=0;
while(x>0)
{
sum+=tree[x];
x-=x&(-x);
}
return sum;
}
ll Get_sum(ll x)
{
return Sum(x,c1)*x-Sum(x,c2);
}
ll Get(ll a,ll b)
{
return Get_sum(b+1)-Get_sum(a);
}

int main()
{
ll q;
ll x;
//freopen("input.txt","r",stdin);
char str[2];
memset(c1,0,sizeof(c1));
memset(c2,0,sizeof(c2));
memset(air,0,sizeof(air));
scanf("%lld %lld",&n,&q);
{
memset(air,0,sizeof(air));
for(int i=1;i<=n;i++)
{
scanf("%lld",&x);
air[i]=air[i-1]+x;

}
while(q--)
{
ll a,b,c;
scanf("%s",str);
if(str[0]=='C')
{
scanf("%lld %lld %lld",&a,&b,&c);
add(c1,a,c);
add(c1,b+1,-c);
add(c2,a,a*c);
add(c2,b+1,-c*(b+1));
}
if(str[0]=='Q')
{
scanf("%lld %lld",&a,&b);

ll ans=Get(a,b);
ans+= air[b]-air[a-1];
cout<<ans<<endl;
}
}
}
return 0;
}