uva 10721

题目链接:uva 10721 - Bar Codes


题目大意:给出n,k和m,用k个1~m的数组成n,问有几种组成方法。


解题思路:简单dp,cnt[i][j]表示用i个数组成j, cnt[i][j] = ∑(1 ≤ t  ≤min(k, j)) cnt[i - 1][t].


#include <stdio.h>
#include <string.h>
#define ll long long
const int N = 105;

ll cnt[N][N];
int n, k, m;

void init() {
	memset(cnt, 0, sizeof(cnt));
	cnt[0][0] = 1;
}

void solve() {
	init();
	for (int i = 1; i <= k; i++) {
		for (int j = 1; j <= n; j++) {
			for (int t = 1;  t <= m && t <= j; t++)
				cnt[i][j] += cnt[i - 1][j - t];
		}
	}	
	printf("%lld
", cnt[k][n]);
}

int main () {
	while (scanf("%d%d%d", &n, &k, &m) == 3) {
		solve();
	}
	return 0;
}


原文地址:https://www.cnblogs.com/suncoolcat/p/3400309.html