uva 10943How do you add?（简单组合）

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 thanN 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

题意：给你n和k，让你求出有k个数的和为n的种类数（可以为0）；很明显的插板法，把n看成n个1，因为可以为0，所以还要加上k个0，即 1 1 1 1 .............0 0 0 0，然后共有n+k-1个空，插入k-1个板，这样的最终结果

代码：

#include <iostream>#include<stdio.h>#include<string.h>#include<stdlib.h>#include<vector>#include<stdlib.h>#include<string>#include<map>#include<set>#include<queue>#include<stack>#include<set>#define inf 0x3f3f3f3f#define eps 1e-5#define max(a,b) (a)>(b)?(a):(b)#define min(a,b) (a)<(b)?(a):(b)#define N 210#define MOD 1000000using namespace std;int c[N+1][N+1];void init(){    int i,j;    c[0][0]=1;    for(i=1;i<=N;i++)    {        c[i][0]=1;        for(j=1;j<N;j++)        {            c[i][j]=(c[i-1][j]%MOD+c[i-1][j-1]%MOD)%MOD;        }        c[i][N]=1;    }}int main(){    int n,k,ans,i;    init();    while(~scanf("%d%d",&n,&k)&&(n||k))    {        printf("%d\n",c[n+k-1][k-1]);    }    return 0;}