How do you add?

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 Nadd 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

译：

[Description]Larry 的数学非常不好，他经常使用计算器。不幸的是，他现在和他一个好朋友被困在一个沙漠岛上。他们正试着通过解决一些好问题来消耗时间，而如果 Larry  不能答出问题，Ryan就要吃掉他，所以他的命运掌握在你的手中！这是一个很简单的问题：给出一个数字 N，用 K 个小于 N 的数加起来为 N，有多少张方法?例如 N = 20, K = 2，就有 21 中方法：0+20 1+19 2+18 3+17 4+16 5+15 ... 18+2 19+1 20+0[Input]对于每组数据一行两个整数 N 和 K直到 N = K = 0 时输入结束[Output]对于每组数据，输出答案对 1,000,000 取模后的结果[Sample Input]20 220 20 0[Sample Output]2121[Hint]1 <= N, K <= 100

/*刚看时觉得比较麻烦但是画表分析下就.......

dp[n][k]

dp[1][1]=1dp[2][1]=1dp[3][1]=1dp[4][1]=1dp[5][1]=1dp[1][2]=2dp[2][2]=3dp[3][2]=4......dp[1][3]=3dp[2][3]=6dp[3][3]=10......dp[1][4]=4...........

把表列出来后会有惊奇的发现

上代码了

#include<iostream>#include<cstdio>#include<cstring>#define mo1000000using namespace std;/*dp[i][j]表示用j个数加起来为i的方法数;dp[i][j]=dp[i-1][j]+dp[i][j-1];边界：dp[i][1]=1;dp[1][j]=j; */int dp[120][120];int n,k;int main(){for(int i=1;i<=100;i++)dp[i][1]=1;for(int i=1;i<=100;i++)dp[1][i]=i;for(int i=2;i<=100;i++)for(int j=2;j<=100;j++){dp[i][j]=(dp[i-1][j]%mo+dp[i][j-1]%mo)%mo;}while(1){scanf("%d%d",&n,&k);if(n==0&&k==0)break;printf("%d\n",dp[n][k]);}return 0;}