UVA10943- How do you add?(隔板法)
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题目链接
题意:求由K个不大于N的非负整数组合而成N的方法数。
思路:隔板法。这个问题可以等价于有N个小球放入K个盒子中,盒子可以为空,问有几种放置的方法。但因为盒子为空,不符合隔板法,所以我们可以先往每个盒子里面放置一个小球,满足每个盒子都不为空的,等分完后,在每个盒子减去一个小球。所以就相当与将N + K - 1个小球分为K - 1份。之后就是组合数学的问题了。
代码:
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;typedef long long ll;const int MOD = 1000000;const int MAXN = 205;int n, k;ll c[MAXN][MAXN];void init() { memset(c, 0, sizeof(c)); for (int i = 0; i < MAXN; i++) c[i][0] = 1; for (int i = 1; i < MAXN; i++) for (int j = 1; j < MAXN; j++) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % MOD;}int main() { init(); while (scanf("%d%d", &n, &k)) { if (n == 0 && k == 0) break; printf("%lld\n", c[n + k - 1][k - 1]); } return 0;}
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