uva 10721 Bar Codes (DP)

Problem D	Bar Codes
Time Limit	1 Second

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 exactlyn 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

#include <iostream>
#include <cstdio>
using namespace std;

const int maxn = 60;
long long int dp[maxn][maxn][maxn][2];
int n , k , m;

void initial(){
for(int i = 0;i < maxn;i++){
for(int j = 0;j < maxn;j++){
for(int l = 0;l < maxn;l++){
dp[i][j][l][0] = 0;
dp[i][j][l][1] = 0;
}
}
}
}

void computing(){
dp[1][1][1][1] = 1;
for(int i = 1;i < n;i++){
for(int j = 1;j <= m;j++){
for(int l = 1;l <= k;l++){
if(dp[i][j][l][0] != 0){
dp[i+1][j+1][l][0] += dp[i][j][l][0];
dp[i+1][1][l+1][1] += dp[i][j][l][0];
}
if(dp[i][j][l][1] != 0){
dp[i+1][j+1][l][1] += dp[i][j][l][1];
dp[i+1][1][l+1][0] += dp[i][j][l][1];
}
}
}
}
long long int sum = 0;
for(int i = 1;i <= m;i++){
sum += dp
[i][k][0];
sum += dp
[i][k][1];
}
cout << sum << endl;
}

int main(){
while(cin >> n >> k >> m){
initial();
computing();
}
return 0;
}

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