Using recursion to solve problems: EOJ2999

Topic:

Description

Given a polynomial (ax+by) k, calculates the coefficients of the xnym term after the expansion of the polynomial.

Input

Line 1th: An integer T (1≤t≤10) is the number of questions.

Next Total T-line. 5 integers per line, respectively, A,b,k,n,m, separated by a space between integers.

0≤k≤1,000,0≤n,m≤k, and n+m=k,0≤a,b≤1,000,000.

Output

For each issue, output a line number of questions (0 start number, format: Case #0: etc.).

Each problem is then output an integer in a row that represents the coefficients (which can be very large, and the value of the output to 10007 ).

Sample Input

3

2 5 290) 130 160

235823 382573 999) 111 888

1 1 3) 1 2

Sample Output

Case #0:

1580

Case #1:

1952

Case #2:

3

#include <stdio.h>#include<stdlib.h>Const intmo=10007;intans[1010][1010];intA,b,k,n,m,i,j,d=0;intcount;voidMain () {scanf ("%d",&count);  while(count--) {scanf ("%d%d%d%d%d",&a,&b,&k,&n,&m); A=a%mo; b=b%mo; ans[0][0]=1;  for(i=1; i<n+1; i++) {ans[i][0]=a*ans[i-1][0]%mo; }         for(j=1; j<m+1; j + +) {ans[0][j]=b*ans[0][j-1]%mo; }         for(i=1; i<n+1; i++){             for(j=1; j<m+1; j + +) {Ans[i][j]= (a*ans[i-1][j]+b*ans[i][j-1])%mo; }} printf ("Case #%d:\n%d\n", d++, Ans[n][m]); }}

At first I wanted to do it with a recursive return but timed out.

intccintKintN) {    intresult=0; if(k==0|| k==N) {result=1; returnresult; }    Else{result= (CC (k,n-1) +CC (K-1, N-1))%10007; returnresult; }}

Using recursion to solve problems: EOJ2999