666: Put the Apples

666: Put the Apples

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Total time limit:

1000ms

Memory Limit:

65536kB

Describe

Put m the same apples on n the same plate, allow some plates to be empty, ask how many different ways? (denoted by k) 5,1,1 and 1,5,1 are the same kind of sub-method.

Input

The first line is the number of test data t (0 <= T <= 20). Each of the following lines contains two integers m and n, separated by a space. 1<=m,n<=10.

Output

For each set of data entered M and N, the corresponding k is output in one line.

Sample input

17 3

Sample output

8

Source

[email protected]

#include <cstdio>using namespacestd;inta[ One][ One],n,m,t;voidinit () {/*Problem-Solving analysis: Set F (m,n) for M apples, n plates of the number of methods, then the first discussion of N, when n>m: there must be n-m a plate forever empty, remove them on the number of apples to be placed no effect.          That is, if (n>m) f (m,n) = f (m,m) when n<=m: Different methods can be divided into two categories: 1, there is at least one plate empty, that is equivalent to f (m,n) = f (m,n-1);        2. All plates have apples, which is equivalent to taking an apple from each plate, without affecting the number of different methods, i.e. f (m,n) = f (m-n,n). The total number of put apples is equal to the sum of the two, that is F (m,n) =f (m,n-1) +f (m-n,n)*/             //A[i][j] means I put an apple on a J-plate.     for(intI=1; i<=Ten; i++) a[0][i]=1, a[i][1]=1;  for(intI=1;i< One; i++)         for(intj=2; j<= One; j + +)            if(i>=j) A[i][j]=a[i][j-1]+a[i-J]            [j]; ElseA[i][j]=a[i][j-1];}intMain () {init (); scanf ("%d",&t);  while(t--) {scanf ("%d%d",&n,&m); printf ("%d\n", A[n][m]); }    return 0;}

666: Put the Apples