UVa 10721 Bar Codes (DP)_acm bar codes-优快云博客

10721 - Bar Codes

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1662

A bar-code symbol consists of alternating dark and light bars, starting with a dark bar on the left. Each bar is a number of units wide. Figure 1 shows a bar-code symbol consisting of 4 bars that extend over 1+2+3+1=7 units.

Figure 1: Bar-code over 7 units with 4 bars

In general, the bar code BC(n,k,m) is the set of all symbols with k bars that together extend over exactly n units, each bar being at most m units wide. For instance, the symbol in Figure 1 belongs to BC(7,4,3) but not to BC(7,4,2). Figure 2 shows all 16 symbols in BC(7,4,3). Each `1' represents a dark unit, each `0' a light unit.

0: 1000100 | 4: 1001110 | 8: 1100100 | 12: 1101110
1: 1000110 | 5: 1011000 | 9: 1100110 | 13: 1110010
2: 1001000 | 6: 1011100 | 10: 1101000 | 14: 1110100
3: 1001100 | 7: 1100010 | 11: 1101100 | 15: 1110110

Figure 2: All symbols of BC(7,4,3)

Input
Each input will contain three positive integers n, k, and m (1 ≤ n, k, m ≤ 50).

Output
For each input print the total number of symbols in BC(n,k,m). Output will fit in 64-bit signed integer.

Sample Input	Output for Sample Input
7 4 3 7 4 2	16 4

思路：我们可以设dp[i][j]表示有i个bar，j个unit的情况数，那么dp[i][j]=sum{dp[i-1][j-k]},1<=k<=M。

完整代码：

/*0.016s*/

#include<cstdio>
#include<cstring>
const int MAXD = 55;

long long dp[MAXD][MAXD];

int main()
{
	int N, K, M, i, j, k;
	while (~scanf("%d%d%d", &N, &K, &M))
	{
		memset(dp, 0, sizeof(dp));
		for (i = 1; i <= M && i <= N; ++i) dp[1][i] = 1;///仅有一个相邻块的情况肯定只有一种
		for (i = 2; i <= K; ++i)
			for (j = i; j <= N; ++j)
				for (k = 1; k < j && k <= M; ++k)
					dp[i][j] += dp[i - 1][j - k];
		printf("%lld\n", dp[K][N]);
	}
	return 0;
}