Number of cycles (M-brute force Solver, play table)

Number of Cycles

Description

we know that in programming, we often need to take into account the complexity of time, especially for the loop part.         For example, if a for (i=1;i<=n;i++) OP appears in the code; Then do N-Times op, if the code appears fori=1;i<=n;         i++) for (j=i+1;j<=n; J + +) OP;         Then do N (n-1)/2 op operation.        Now you are given an M-layer for loop operation, and each time the start value of the variable in for is the starting value of the previous variable +1 (the first variable starts with a value of 1), the terminating value is an input n, and the last OP has a total amount of computation.         

Input

There are T group case,t<=10000.       Each case has two integers m and n,0<m<=2000,0<n<=2000.                 

Output

for each case, output a value that represents the total amount of computation, perhaps the number is large, then you only need to output the remainder of the 1007 left.         

Sample Input

2 1 3) 2 3                

Sample Output

3 3Test Instructions:calculates the number of runs of a variable in a for loop. Analysis:1. Math problems, permutations and combinations. Formula: C (n,m) =c (n-1,m) +c (n-1,m-1)2. You need to select the number of m from the N number and increment the loop

Code:

1#include <cstdio>2#include <iostream>3 using namespacestd;4 5 intt[10000],m[ -],n[ -];6 inta[2005][2005];7 8 intMain ()9 {Ten     intT; Onescanf"%d",&T); Aa[0][0]=1;  -  -      for(intI=1; i<= -; i++)//Play Table the     {           -a[i][0]=1; -          for(intj=1; j<= -; j + +) -A[i][j]= (a[i-1][j]+a[i-1][j-1])%1007; +     } -  +      while(t--) A     {     at         intM,n; -scanf"%d%d",&m,&N);  -printf"%d\n", A[n][m]); -     } - return 0; -}

View Code

Number of cycles (M-brute force Solver, play table)