Paint Chain

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2741    Accepted Submission(s): 995

Problem Description

Aekdycoin and abcdxyzk are playing a game. They get a circle chain with some beads. Initially none of the beads is painted. They take turns to paint the chain. In Each turn one player must paint a unpainted beads. Whoever is unable to paint in his turn lose the game. Aekdycoin will take the first move.

Now, they thought this game is too simple, and they want to change some rules. In each turn one player must select a certain number of consecutive unpainted beads to paint. The other rules is The same as the original. Who will win under the rules ?You may assume that both of them are so clever.

 

Input

First line contains T, the number of test cases. Following T line contain 2 integer N, M, indicate the chain has N beads, and each turn one player must paint M consecutive beads. (1 <= N, M <= 1000)

 

Output

For each case, print "Case #idx: " first where idx is the case number start from 1, and the name of the winner.

 

Sample Input

	

		2 
3 1 
4 2
	

 

Sample Output

	

		Case #1: aekdycoin 
Case #2: abcdxyzk
	

 

Author

jayi

 

Source

 

#include <cstdlib>
#include <cmath>
#include <cstdio>
#include <memory.h>

using namespace std;

int result[1001];
bool g[1001];

void compute(int count, int paints) {
    int i;

    int subCount = count - paints;

    if (subCount < 0) {
        result[count] = 0;
        return;
    }

    memset(g, false, sizeof(g));

    for (i = 0; i <= subCount && i <= subCount - i; ++i) {
        g[result[i] ^ result[subCount - i]] = true;
    }

    for (i = 0; ;++i) {
        if (g[i] == false) {
            result[count] = i;
            break;
        }
    }
}

int main()
{
    int i, j;

    int num;
    int beads;
    int paints;

    scanf("%d", &num);
    for (i = 0; i < num; ++i) {
        memset(result, -1, sizeof(result));
        scanf("%d %d", &beads, &paints);

        if (paints > beads) {
            printf("Case #%d: abcdxyzk\n", i + 1);
            continue;
        }

        beads -= paints;

        for (j = 0; j <= beads; ++j) {
            compute(j, paints);
        }

        if (result[beads] == 0) {
            printf("Case #%d: aekdycoin\n", i + 1);
        } else {
            printf("Case #%d: abcdxyzk\n", i + 1);
        }
    }

    return 0;
}