Hrbust 2325 Astonishing Combinatorial Number【前缀和+暴力】

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本文介绍了一种通过预处理杨辉三角求特定条件下组合数Ci,j%k==0的个数的方法。对于给定的n、m和k，文章详细阐述了如何高效地计算满足条件的组合数对(i,j)的数量。

Astonishing Combinatorial Number

Time Limit: 8000 MS

Memory Limit: 512000 K

Total Submit: 58(12 users)

Total Accepted: 15(10 users)

Rating:

Special Judge: No

Description

    The combinatorial number CmnCnm in
 combinatorics, where it gives the number of the ways, disregarding order, 
the m objects can be chosen from among n objects; more formally, the number of m-element subsets (or m-combinations)
 of an n-element set.
    Now, there's a simple problem about this number:

    For three given number n, m and k, how many pairs of (i, j) will make the equation A.a be true. (0 <= i <= n, 0 <= j <= min(i, m))

    Equation A.1:

                    Cjimodk=0Cijmodk=0

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases.
For each test case:

The first line contains two integers  t(1<=t<=10^4) and k(1<=k<=30).

Each of the next t lines contains two integers  n(1<=n<=2000) and m(1<=m<=2000) .

Output

Each test case should contain t lines. Each line has an integer indicate the answer of the problem.

Sample Input

Sample Output

1
0
7

Hint

In first test case, only C12=2C21=2 is
 multiple of 2.

Source

"尚学堂杯"哈尔滨理工大学第七届程序设计竞赛

题目大意：

给你t个查询，每个查询询问Ci,j%k==0的个数（i<=n,j<=min(m,i)）。

思路：

我们知道，Ci,j的求法有很多，这里数据范围并不大，而且模数不定，所以我们肯定要用求杨辉三角的方法来求Ci,j的。

那么对于每个k，我们预处理出来C【2000】【2000】表示Ci,j%k的值，然后我们维护一个前缀和，O（t*n）查询即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int c[2005][2005];
int sum[2005][2005];
void init(int z)
{
    memset(sum,0,sizeof(sum));
    memset(c,0,sizeof(c));
    c[0][0]=1%z;
    c[1][0]=1%z;
    c[1][1]=1%z;
    for(int i=2;i<=2002;i++)
    {
        c[i][0]=1%z;
        for(int j=1;j<=i;j++)
        {
            c[i][j]=(c[i-1][j]+c[i-1][j-1])%z;
        }
    }
    for(int i=1;i<=2002;i++)
    {
        for(int j=1;j<=2002;j++)
        {
            int tmp=0;
            if(c[i][j]==0)tmp=1;
            sum[i][j]=sum[i][j-1]+tmp;
        }
    }
}
int main()
{
    int t;
    while(~scanf("%d",&t))
    {
        while(t--)
        {
            int q,k;
            scanf("%d%d",&q,&k);
            init(k);
            while(q--)
            {
                int n,m;
                scanf("%d%d",&n,&m);
                int output=0;
                for(int i=1;i<=n;i++)
                {
                    output+=sum[i][min(i,m)];
                }
                printf("%d\n",output);
            }
        }
    }
}