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;

}

View Code