UVA 10943 - How do you add?

最新推荐文章于 2018-08-21 11:23:23 发布

转载最新推荐文章于 2018-08-21 11:23:23 发布 · 117 阅读

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原文链接：http://www.cnblogs.com/toyking/p/3797369.html

本文介绍了一个数学问题，即如何计算将一个数N拆分为K个正整数之和的不同方法的数量，并提供了一段C++代码实现。该问题通过动态规划解决，代码中详细展示了初始化过程及核心计算逻辑。

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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 Nadd 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：N拆成K个正整数的和，有多少种方法？

dp[N][K] 记录和为N，由K个数组成

则有dp[N][K]=sum{ dp[N-1][k-i] }

#include <iostream>
using namespace std;

const int mod=1000000;
int dp[110][110];

void ini(){
	dp[0][0]=1;
	for(int k=1;k<=100;k++){
		for(int sum=0;sum<=100;sum++){
			for(int i=0;i<=100-sum;i++){
				dp[sum+i][k]=(dp[sum+i][k]+dp[sum][k-1])%mod;
			}
		}
	}
}

int main(){
	ini();
	int n,k;
	while(cin>>n>>k && (n||k)){
		cout<<dp[n][k]<<endl;
	}
	return 0;
}

转载于:https://www.cnblogs.com/toyking/p/3797369.html