UVA 10943 How do you add?

Problem A: How do you add?

Larry is very bad at math - he usually uses a calculator, which worked well throughout college. Unforunately, he is now struck in a deserted island with his good buddy Ryan after a snowboarding accident. They're now trying to spend some time figuring out some good problems, and Ryan will eat Larry if he cannot answer, so his 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 = 20 and K = 2, there are 21 ways: 
0+20 
1+19 
2+18 
3+17 
4+16 
5+15 
... 
18+2 
19+1 
20+0 

Input

Each line will contain a pair of numbers N and K. N and K will both be an integer from 1 to 100, inclusive. The input will 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 on a single line. 

Sample Input

20 220 20 0

Sample Output

2121

这题主要是找规律，因为需要找一个递推公式，可以看到下面的

n k 1 2 3 4 5 …… 100
1    1 2 3 4 5…… 100
2    1 3 6 10 15……
3    1 4 10 20 35 ……
4    1 5 15 35 70 ……
分析后就可发现有   a[i][j]=a[i-1][j]+a[i][j-1]，事先要对数组进行预处理，就是a[0][i](0<=i<=99)=i,a[i][0](0<=i<=99)=1，这是我们可以知道的，也是递推的前提条件，然后再对结果进行取模就好了
#include<iostream># define mod 1000000using namespace std;int main(){int n,k,i,j,a[100][100];for(i=0;i<100;i++){a[i][0]=1;a[0][i]=i+1;}for(i=1;i<100;i++)for(j=1;j<100;j++)a[i][j]=(a[i][j-1]+a[i-1][j])%mod;while(cin>>n>>k,n&&k){cout<<a[n-1][k-1]<<endl;}return 0;}