UVA 10943 HOW do you add?

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本文介绍了一个有趣的数学问题：如何计算K个数相加等于N的所有可能组合数量。使用动态规划解决这一问题，提供了完整的C++代码实现。

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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 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
**提议概述：给你一个数N和K，让你将N拆分为K个数，并让你求出有多少种拆分的方法？
思路：这道题看到的时候，开始感觉挺忙然，但你会发现这道题用dp是不错的选择；
核心：dp[N][K]=dp[i][k-1] (0<=i<=N)**
(http://acm.hust.edu.cn/vjudge/contest/121559#problem/B)

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int dp[105][105];
void inti()
{
    memset(dp,0,sizeof(dp));
    int i,j,k;
    for(i=0;i<105;i++)
        dp[0][i]=1;
    for(k=1;k<105;k++)
    {
        for(j=1;j<105;j++)
        {
            for(i=0;i<=j;i++)
                dp[j][k]=(dp[j][k]+dp[i][k-1])%1000000;
        }

    }
}
int main()
{
    int n,k;
    inti();
    while (~scanf("%d%d",&n,&k))
    {
        if(n==0&&k==0)
            break;
        printf("%d\n",dp[n][k]);
    }
    return 0;
}