UVA 10943 - How do you add?(dp)

最新推荐文章于 2024-07-27 00:32:31 发布

lab104_yifan

最新推荐文章于 2024-07-27 00:32:31 发布

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本文链接：https://blog.youkuaiyun.com/accelerator_/article/details/12874871

本文讨论了如何计算给定数值N和K时，小于N的K个数相加得到N的所有可能组合数量。通过动态规划方法，提供了解决这类问题的高效算法，并附上实现代码。

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

题意：给定n，k，求出由k个数字加出n的方法种数。

思路: dp。 i表示和，j表示种数， dp[i][j] = dp[i - 1][j] + dp[i][j - 1],

代码：

#include <stdio.h>
#include <string.h>

long long n, k, dp[105][105];

int main() {
    for (int i = 1; i <= 100; i ++) {
	dp[i][1] = 1;
	dp[1][i] = i;
    }
    for (int i = 2; i <= 100; i ++) {
	for (int j = 2; j <= 100; j ++) {
	    dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % 1000000;
	}
    }
    while (~scanf("%lld%lld", &n, &k) && n || k) {
	printf("%lld\n", dp[n][k]);
    }
    return 0;
}