/*
# 问题分析
题设给定三个数值ability, X, Y, 以及三个关系式(三元一次不等式组). 要求根据X, Y计算出满足三个关系式的最大解.
然后根据ability跟不等式组解出来的三个值的大小关系输出对应的关键字.
如果不等式组没有解, 则输出"No Solution".
 
# 完整描述步骤
1. 获取输入: 你的能力值, 倍数X, 倍数Y
2. 从99到10遍历甲的能力值:
    - 计算 乙的能力值 = 甲的能力值 % 10 * 10 + 甲的能力值 % 10
    - 计算 丙的能力值 = 绝对值(甲的能力值 - 乙的能力值) 除以 X
    - 检查甲乙丙能力值是否满足关系式:
        - 如果 乙的能力值 == 丙的能力值 * Y:
            - 输出甲的能力值
            - 输出你的能力值与甲的能力值之间的关系
            - 输出你的能力值与乙的能力值之间的关系
            - 输出你的能力值与丙的能力值之间的关系
            - 结束程度
3. 输出"No Solution"
 
# 伪代码描述
1. get input: your_ability, X, Y
2. for ability_A in range(99, 9, -1):
    - calculate ability_B = ability_A % 10 * 10 + ability_A / 10;
    - calculate ability_C = abs(ability_B - ability_A) / X;
    - if ability_B == ability_C * Y:
        - print(ability_A);
        - print(relation between your_ability and ability_A);
        - print(relation between your_ability and ability_B);
        - print(relation between your_ability and ability_C);
        - end procedure;
 
3. print("No Solution")
 
*/
 
 
#include <math.h>
#include <stdio.h>
 
void print(double your_ability, double others_ability) {
    if (your_ability == others_ability) {
        printf(" Ping");
    } else if (your_ability < others_ability) {
        printf(" Cong");
    } else {
        printf(" Gai");
    }
}
 
int main() {
    int your_ability, X, Y;
    scanf("%d %d %d", &your_ability, &X, &Y);
    for (int ability_A = 99; ability_A >= 10; ability_A--) {
        int ability_B = ability_A % 10 * 10 + ability_A / 10;
        double ability_C = abs(ability_B - ability_A) * 1.0 / X;
        if (ability_B == ability_C * Y) {
            printf("%d", ability_A);
            print(your_ability, ability_A);
            print(your_ability, ability_B);
            print(your_ability, ability_C);
            return 0;
        }
    }
 
    printf("No Solution");
    return 0;
}