How do you add?------组合数学，隔板

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 2
20 2
0 0
Sample Output
21

21

题目大意：大小为n的数分成m组有多少种

题意：给你n和k，让n分成K组（可以为0）；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.

C(n-k+1,k-1)

隔板法........:

https://baike.baidu.com/item/%E9%9A%94%E6%9D%BF%E6%B3%95/3902458?fr=aladdin

 

#include
#include
#define mod 1000000
using namespace std;
typedef long long ll;
int main()
{
    ll n,k,i,j,c[300][300];
    c[0][0]=1;c[1][0]=1;c[1][1]=1;
    for(i=2;i<200;i++)
    {
        c[i][0]=1;c[i][i]=1;
        for(j=1;j

 2.???

用d（i，j）表示i拆成j个数和的种类数

d（i，j）=d（i-1，j）+d（i，j-1）

边界：d（i，1）=1

#include 
#include 
#define mod 1000000
int main()
{
    int p[110][110],n,k;
    memset(p,0,sizeof(p));
    for(int i=0;i<=100;i++)
    {
        p[i][1]=1;
        for(int j=2;j<=100;j++)
        {
            p[i][j]=(p[i-1][j]%mod+p[i][j-1]%mod)%mod;
        }
    }
    while(scanf("%d%d",&n,&k)==2)
    {
        if(!n&&!k)break;
        printf("%d\n",p[n][k]);
    }
    return 0;
}