uva 10943 How do you add? (DP)

本文介绍了一个有趣的数学问题:给定一个数N,求K个小于N的数相加等于N的所有组合数量。通过动态规划算法解决该问题,并提供了一种高效求解方案,最后输出答案对1,000,000取模的结果。

Problem E : How do you add?

From:UVA, 10943

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 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
 
#include <iostream>
#include <cstdio>
using namespace std;

const int maxn = 110;
int dp[maxn][maxn];
int N , K;

void initial(){
	for(int i = 0;i <= 100;i++){
		dp[i][1] = 1;
		dp[0][i] = 1;
	}
	for(int i = 1;i <= 100;i++){
		for(int k = 2;k <= 100;k++){
			for(int j = 0;j <= i;j++){
				dp[i][k] = (dp[i][k]+dp[j][k-1])%1000000;
			}
		}
	}
}

int main(){
	initial();
	while(cin >> N >> K && (N||K)){
		cout << dp[N][K] << endl;
	}
	return 0;
}


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