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

;

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
[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
[k]);
}
return 0;
}