UVA 10943 How does you add?

Original question:
Larry is very bad at Math-he usually uses a calculator, which
worked well throughout college. Unforunately, he is now struck in
A deserted and his good buddy Ryan after a snowboarding
Accident.
They ' re now trying to spend some time figuring out some good
Problems, and Ryan would eat Larry if he cannot answer, so he fate
is up to you!
It's a very simple problem-given a number N, how many ways
Can K numbers less than n add up to n?
For example, for N = 2 and K = +, there is ways:
0+20
1+19
2+18
3+17
4+16
5+15
...
18+2
19+1
20+0
Input
Each line would contain a pair of numbers n and K. N and K would both is an integer from 1 to 100,
Inclusive. The input would terminate on 2 0 ' s.
Output
Since Larry is only interested in the last few digits of the answer, for each pair of numbers N and K,
Print a single number mod 1,000,000 to a single line.
Sample Input
20 2
20 2
0 0
Sample Output
21st
21st
English:
Give you two numbers, N and K, and ask you how many ways to make up n with the number of K, in which the number of components is in order. For example, when n=20,k=2
0+20
1+19
2+18
3+17
4+16
5+15
...
18+2
19+1
20+0
There are 21 kinds

#include <bits/stdc++.h>
using namespace std;
int dp[101][101];
const int mod=1000000;
int main ()
{
    Ios::sync_with_stdio (false);
    int n,k;
    Memset (Dp,0,sizeof (DP));
    for (int i=1;i<=100;i++)
        dp[0][i]=dp[i][1]=1;
    for (int i=1;i<=100;i++) for
        (int j=1;j<=100;j++)
            dp[i][j]= (dp[i-1][j]%mod+dp[i][j-1]%mod)%mod;
    while (cin>>n>>k,n+k)
        cout<<dp[n][k]<<endl;
    return 0;
}

Answer:
A simple combination of numbers or dynamic programming is a topic that is pushed together.
Set state Dp[n][k] means useful K number composition n How many ways, then the formula is
DP[N][K]=DP[N-1][K]+DP[N][K-1]
For example, n=5,k=2 can be added by n=4,k=2 plus n=5,k=1 and get,
N=4,k=2 when the
0 4
1 3
2 2
3 1
4 0
Then the number on the right column, plus 1, gets 5.
But there is a lack of a
5 0
Plus, the equivalent of Dp[5][1] adds another 0

You can also consider the idea of putting the ball inside the box.