NKOI 1321--数列操作问题(裸BIT)

数列操作问题

Time Limit:10000MS  Memory Limit:65536K
Total Submit:276 Accepted:149 
Case Time Limit:1000MS

Description

假设有一列数{Ai}(1≤i≤n)，支持如下两种操作： 
将Ak的值加D。（k, D是输入的数） 
输出As+As+1+…+At。（s, t都是输入的数，S≤T）

Input

第一行一个整数n， 
第二行为n个整数，表示{Ai}的初始值≤10000。 
第三行为一个整数m，表示操作数 
下接m行，每行描述一个操作，有如下两种情况： 
ADD k d (表示将Ak加d，1<=k<=n，d为数,d的绝对值不超过10000) 
SUM s t （表示输出As+…+At） 

Output

对于每一个SUM提问，输出结果

Sample Input

 

5 
1 2 3 2 4 
5 
SUM 1 2
SUM 1 5
ADD 1 2
SUM 1 2
SUM 1 5

 

Sample Output

 

3 
12 
5 
14 

 

Hint

M,N<=100000

Source

解题思路：

　　模板题

代码：

# include<cstdio>
# include<iostream>
# include<fstream>
# include<algorithm>
# include<functional>
# include<cstring>
# include<string>
# include<cstdlib>
# include<iomanip>
# include<numeric>
# include<cctype>
# include<cmath>
# include<ctime>
# include<queue>
# include<stack>
# include<list>
# include<set>
# include<map>

using namespace std;

const double PI=4.0*atan(1.0);

typedef long long LL;
typedef unsigned long long ULL;

# define inf 999999999
# define MAX 100000+4

int n;
int a[MAX];
int tree[MAX];//树状数组
char s[23];

int read ( int pos )
{
    int temp = 0;
    for ( int j = pos;j;j-=j&(-j) )
    {
        temp+=tree[j];
    }
    return temp;
}

void update( int pos,int val )
{
    for ( int j = pos;j <= n;j+=j&(-j) )
    {
        tree[j]+=val;
    }
}

int main(void)
{
    char c;
    scanf("%d",&n);
    for ( int i = 1;i <= n;i++ )
    {
        scanf("%d",&a[i]);
    }
    for ( int i = 1;i <= n;i++ )
    {
        update(i,a[i]);
    }
    int m;scanf("%d\n",&m);
    for ( int i = 0;i < m;i++ )
    {
        int x,y;
        int ans1 = 0;
        int ans2 = 0;
        scanf("%s",s);
        scanf("%d",&x);
        scanf("%d",&y);
        //cout<<x<<" "<<y<<endl;
        if ( s[0] == 'S' )
        {
            ans1 = read(y);
            ans2 = read(x-1);
            printf("%d\n",ans1-ans2);
        }
        else if ( s[0] == 'A' )
        {
            update(x,y);
        }
    }

    return 0;
}