HDU 4267 A Simple Problem with Integers 第37届ACM/ICPC长春赛区网络赛1001题 （树状数组）

A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 659    Accepted Submission(s): 253

Problem Description

Let A1, A2, ... , AN be N elements. You need to deal with two kinds of operations. One type of operation is to add a given number to a few numbers in a given interval. The other is to query the value of some element.

 

 

Input

There are a lot of test cases.
The first line contains an integer N. (1 <= N <= 50000)
The second line contains N numbers which are the initial values of A1, A2, ... , AN. (-10,000,000 <= the initial value of Ai <= 10,000,000)
The third line contains an integer Q. (1 <= Q <= 50000)
Each of the following Q lines represents an operation.
"1 a b k c" means adding c to each of Ai which satisfies a <= i <= b and (i - a) % k == 0. (1 <= a <= b <= N, 1 <= k <= 10, -1,000 <= c <= 1,000)
"2 a" means querying the value of Aa. (1 <= a <= N)

 

 

Output

For each test case, output several lines to answer all query operations.

 

 

Sample Input

4 1 1 1 1 14 2 1 2 2 2 3 2 4 1 2 3 1 2 2 1 2 2 2 3 2 4 1 1 4 2 1 2 1 2 2 2 3 2 4

 

 

Sample Output

1 1 1 1 1 3 3 1 2 3 4 1

 

 

Source

2012 ACM/ICPC Asia Regional Changchun Online

 

 

Recommend

liuyiding

 

 

 

长春网络赛第一题。

一看这题没有什么思路，后来发现k比较小。

马上想到建很多树状数组可以解决。

写好之后交上去就1A了。

做得比较顺手的题目了。

#include<stdio.h>



#include<queue>

#include<iostream>

#include<algorithm>

#include<string.h>

const int MAXN=50020;





int c[12][12][MAXN];

int n;



int lowbit(int x)

{

    return x&(-x);

}



void update(int t1,int t2,int i,int val)

{

    while(i<=n)

    {

        c[t1][t2][i]+=val;

        i+=lowbit(i);

    }



}

int sum(int t1,int t2,int i)

{



    int s=0;

    while(i>0)

    {

        s+=c[t1][t2][i];

        i-=lowbit(i);

    }

    return s;

}

int num[MAXN];

int main()

{

    //freopen("in.txt","r",stdin);

    //freopen("out.txt","w",stdout);



    int m;

    while(scanf("%d",&n)!=EOF)

    {

        for(int i=0;i<n;i++)scanf("%d",&num[i]);

        for(int i=0;i<12;i++)

          for(int j=0;j<12;j++)

            for(int k=0;k<MAXN;k++)

              c[i][j][k]=0;

        scanf("%d",&m);

        int a,b,k,q;

        int t;

        while(m--)

        {

            scanf("%d",&t);

            if(t==1)

            {

                scanf("%d%d%d%d",&a,&b,&k,&q);

                a--;

                b--;

                int num=(b-a)/k;

                int s=a%k;

                update(k,s,a/k+1,q);

                update(k,s,a/k+num+2,-q);

            }

            else

            {

                scanf("%d",&a);

                a--;

                int ss=num[a];

                for(int i=1;i<=10;i++)

                {

                    ss+=sum(i,a%i,a/i+1);

                }

                printf("%d\n",ss);

            }

        }

    }

    return 0;

}