Description
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
Resume
将N分解成K个非负整数之和的方案数。
Analysis
- 思路一：
高中组合数问题，通过隔板法可得答案为 ({C_{n+k-1}}^n) 种。 - 思路二：
动态规划，转移方程为$$d[i][j] = d[i-1][j] + d[i][j-1] $$其中 (d[i][j]) 表示将(i)拆分成(j)个数的方案总数。
Code(Measure One)
//////////////////////////////////////////////////////////////////////
//Target: UVa 10943 - How do you add?
//@Author: Pisceskkk
//Date: 2019-2-16
//////////////////////////////////////////////////////////////////////
#include<cstdio>
#define N 220
#define mod 1000000
#define ll long long
using namespace std;
int n,k,f[N][N];
int dfs(int a,int b){
if(a < b)return 0;
if(b == 0)return 1;
if(a == 0)return 1;
if(f[a][b])return f[a][b];
return f[a][b] = (dfs(a-1,b-1)+dfs(a-1,b))%mod;
}
int main(){
while(1){
scanf("%d %d",&n,&k);
if(!n && !k){
break;
}
printf("%d
",dfs(n+k-1,n));
}
return 0;
}