cogs 1316. Sequence Procedure B modified single-point interval query

1316. The number of columns Procedure B

★★ Input file: shulieb.in   Output file: shulieb.out   a simple comparison of
the time limit: 1 s memory limit: 128 MB

【Problem Description】

Suppose there is a size of  n- ( n- ≤ 100000 )  integer sequence  A , supports the following two operations:

1.  A I , A I + . 1 , ... , A J  values were increased  d

2. Query  A I  value

The operational requirements for proper operation, and outputs the result.

[Input Format]

A first line of the input file integer  n- ,

The second line  n  integer representing the number of columns  A  initial value items.

A third row integer  m  , operands. Then the  m  rows, each row describes an operation, the following two situations:

ADD ijd (the  A I , A I + . 1 , ... , A J ( . 1 ≤ I , J ≤ n- )  values were increased an integer  D )

QUERY s (representing a query  A S  value)

[Output format]

For each inquiry, the results of the query output.

[Sample input]

4
1 4 2 3
3
QUERY 1
ADD 2 2 50
QUERY 2

[Sample output]

. 1 
54 is 


the practice of a: Difference Array storage arrays in a tree 
code is as follows

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#define maxn 100005
using namespace std;
int n,m;
int a[maxn],sum[maxn];
int lowbit(int x){ return x&(-x); }
#define ll long long
void Add(int x,int d){//存的是差分数组 
    while(x<=n){sum[x]+=d;x+=lowbit(x); }
}
long long Sum(int x){
    long long ret=0;
    while(x>0){ ret+=sum[x];x-=lowbit(x);}
    return ret;
}
int main()
{
    freopen("shulieb.in","r",stdin);freopen("shulieb.out","w",stdout);
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    scanf("%d",&m);
    for(int i=1;i<=m;i++){
        string s;cin>>s;
        if(s[0]=='Q'){
            int x;scanf("%d",&x);
            printf("%lld\n",Sum(x)+a[x]);
        }
        else{
            int l,r,x;scanf("%d%d%d",&l,&r,&x);
            Add(l,x);Add(r+1,-x);
        }
            
    }
    
    return 0;
 }