UVa 10721 Bar codes (DP)

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) are the set of all symbols with k bars, together extend over exactly n units, each B Ar 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 symbols in BC (7,4,3). Each ' 1 ' represents a dark unit, each ' 0 ' a light unit.

0:1.0001 million | 4:1,001,110 | 8:1.1001 million | 12:1,101,110

1:1,000,110 | 5:1.011 million | 9:1,100,110 | 13:1,110,010

2:1.001 million | 6:1.0111 million | 10:1.101 million | 14:1.1101 million

3:1.0011 million | 7:1,100,010 | 11:1.1011 million | 15:1,110,110

Figure 2:all Symbols of BC (7,4,3)

Input

Each input would 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'll fit in 64-bit signed integer.

Train of thought: we can set DP[I][J] to indicate the number of cases with the I bar,j unit, then dp[i][j]=sum{dp[i-1][j-k]},1<=k<=m.

See more highlights of this column: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/

Complete code:

01./*0.016s*/
.  
#include <cstdio>  
#include <cstring>  
05.const int maxd =;  
07.long long dp[maxd][maxd];  
09.int Main ()  
10.{    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;///Only one contiguous block must be only one  
.        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;  
23.}