## Algorithm

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. 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 7 4 3 7 4 2

Sample Output 16 4

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

`````` #include <bits/stdc++.h>
using namespace std;

int n,k,m;
long long memo[51][51];
long long solve(int remLength, int remK){
if(remLength == 0 && remK == 0) return 1;
else if(remK == 0 || remLength <=0) return 0;
if(memo[remLength][remK] != -1) return memo[remLength][remK];
long long res = 0;
for(int i=1;i < =m;i++)
res += solve(remLength-i, remK-1);
return memo[remLength][remK] = res;
}

int main() {
while(scanf("%d %d %d",&n,&k,&m) != EOF){
memset(memo,-1,sizeof memo);
printf("%lld\n", solve(n,k)>;
}
}
``````
Copy The Code &

Input

cmd
7 4 3
7 4 2

Output

cmd
16
4