Noip2016day2 Combinatorial number problem problem

Title Description

The number of combinations represents the number of programs that select m items from n items. For example, select two items from three items (1,3), (2,3), and three options to choose from. Based on the definition of combinatorial number, we can give a general formula for calculating the number of combinations:

where n! = 1x2x Xn

Shallot would like to know if given n,m and K, for all 0 <= i <= n,0 <= j <= min (i,m) How many pairs (i,j) satisfy is a multiple of K.

Input/output format

Input format:

The first line has two integer t,k, where T represents the total number of test data for the test point, and the meaning of K is "problem description".

Next T line two integers per line n,m, where the meaning of n,m see "Problem description".

Output format:

T line, one integer per line represents the answer.

Input/Output sample

Input Sample # # :

1 2

7 ·

Output Example # # :

1

Input Example # # :

2 5

4 5

6 7

Output Example # # :

0

7

Description

"Sample 1 description"

In all possible cases, there is only a multiple of 2.

"Subtasks"

Ideas

This is a number proposition.

First, we need to know the general recursive formula of combinatorial number, and its recurrence formula is the same as the Yang Hui Triangle.

C[I][J]=C[I-1][J-1]+C[I-1][J]

(Explanation: C[i][j] is the number of options for selecting J pieces from the I item. If item I is not selected, the number of programmes becomes C[I-1][J], if Item I is selected, the number of programmes becomes c[i-1][j-1], the total programme number is the sum of the number of programmes in two cases)

In order not to explode long long, each time to find C[i][j] after the first model K

In order to save time, the two-dimensional summation, and finally directly find the answer

#include <iostream>using namespacestd;intn,m,t,k;intc[2001][2001],s[2001][2001];voidGet_c () { for(intI=0; i<= -; i++)    {         for(intj=0; j<=i;j++)        {            if(i==0&&j==0) c[i][j]=1%K; Else{C[i][j]= (c[i-1][j]+c[i-1][j-1])%K; }        }    }}voidget_s () {if(c[0][0]==0(s:0][0]=1;  for(intI=0; i<= -; i++)    {         for(intj=0; j<=i;j++)        {            if(i==0&&j==0)Continue; Else            {                if((i==0&AMP;&AMP;J) | | (i==j)) {S[i][j]=s[i][j-1]; if(c[i][j]==0) s[i][j]++; }                Else if(i&&j==0) {S[i][j]=s[i-1][j]; if(c[i][j]==0) s[i][j]++; }                Else if(i&&j) {S[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]; if(c[i][j]==0) s[i][j]++; }            }        }    }}intMain () {CIN>>t>>K;    Get_c ();    get_s ();  while(t--) {cin>>n>>m; cout<<s[n][min (n,m)]<<Endl; }}

Noip2016day2 Combinatorial number problem problem