A Simple Problem with Integers_树状数组

Posted 2020-08-21 阿宝的锅锅

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

 

 

【题意】给出n个数，再给出m个操作

1.op==1,输入四个数据a,b,k,c将区间[a,b]中的数i满足(i-a)%k  == 0加上c.

2.op==2,输入一个数y,输出序列中第y个数的值。

被这题虐哭，（毕竟太弱了~~）关键就是建立多个树状数组，然而我对树状数组理解还是不行啊！

sum[x][k][x%k]代表x对k取余的值，然后每次更新树状数组的时候只需要更新update(a,.....) 与update(b+1,.....);

参考资料：http://blog.csdn.net/yeguxin/article/details/47999833

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int N=50000+10;
int aa[N];
int n,m;
int sum[N][12][12];//开稍大一点就会MLE

int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int k,int mod,int v)
{
    while(x<=n)
    {
        sum[x][k][mod]+=v;
        x+=lowbit(x);
    }
}
int query(int x,int y)
{
    int res=0;
    while(x)
    {
        for(int i=1;i<=10;i++)
        {
            res+=sum[x][i][y%i];
        }
        x-=lowbit(x);
    }
    return res;
}
int main()
{
    while(~scanf("%d",&n))
    {
        memset(sum,0,sizeof(sum));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&aa[i]);
        }
        scanf("%d",&m);
        int op,a,b,k,c;
        while(m--)
        {
            scanf("%d",&op);
            if(op==2)
            {
                scanf("%d",&k);
                int ans=query(k,k);
                printf("%d\n",ans+aa[k]);
            }
            else if(op==1)
            {
                scanf("%d%d%d%d",&a,&b,&k,&c);
                int kk=(b-a)/k;
                update(a,k,a%k,c);
                update(b+1,k,a%k,-c);
            }
        }
    }
    return 0;
}