uva 10943 - How do you add?(dp)

题目连接：uva 10943 - How do you add?

题目大意：给出n和k，用k个数组成加法算式，使得最后的和为n，求有多少种组合方式。

解题思路：一开始不知道0也算，即比如说 1  100， 可以是0 + 0 + 1，所以说答案是100。num[i][j] = ∑ num[k][j - 1](0 ≤ k ≤ i)。

#include <stdio.h>#include <string.h>const int N = 105;const int tmp = 1000000;int num[N][N];void init() {memset(num, 0, sizeof(num));for (int i = 0; i <= 100; i++) num[0][i] = 1;for (int i = 1; i <= 100; i++)for (int j = 1; j <= 100; j++)for (int k = 0; k <= i; k++)num[i][j] = (num[i][j] + num[k][j - 1]) % tmp;}int main () {init();int n, k;while(scanf("%d%d", &n, &k), n || k) {printf("%d\n", num[n][k]);}return 0;}

还看到另外一种更简单的方法，num[i][j] = num[i - 1][j] + num[i][j - 1]。不过不是很清楚公式怎么得到的，求大神指教。

#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;}