Algorithm
After Fox Ciel won an onsite round of a programming contest, she took a bus to return to her castle. The fee of the bus was 220 yen. She met Rabbit Hanako in the bus. They decided to play the following game because they got bored in the bus.
- Initially, there is a pile that contains x 100-yen coins and y 10-yen coins.
- They take turns alternatively. Ciel takes the first turn.
- In each turn, they must take exactly 220 yen from the pile. In Ciel's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 100-yen coins. In Hanako's turn, if there are multiple ways to take 220 yen, she will choose the way that contains the maximal number of 10-yen coins.
- If Ciel or Hanako can't take exactly 220 yen from the pile, she loses.
Determine the winner of the game.
The first line contains two integers x (0 ≤ x ≤ 106) and y (0 ≤ y ≤ 106), separated by a single space.
If Ciel wins, print "Ciel". Otherwise, print "Hanako".
2 2
Ciel
3 22
Hanako
In the first turn (Ciel's turn), she will choose 2 100-yen coins and 2 10-yen coins. In the second turn (Hanako's turn), she will choose 1 100-yen coin and 12 10-yen coins. In the third turn (Ciel's turn), she can't pay exactly 220 yen, so Ciel will lose.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <cstdio>
int main(){
long hundreds(0), tens(0); scanf("%ld %ld", &hundreds, &tens);
bool loser(0);
while(true){
//Ciel
if(hundreds >= 2 && tens >= 2){hundreds -= 2; tens -= 2;}
else if(hundreds >= 1 && tens >= 12){hundreds -= 1; tens -= 12;}
else if(tens >= 22){tens -= 22;}
else{loser = 1; break;}
//Hanako
if(tens >= 22){tens -= 22;}
else if(tens >= 12 && hundreds >= 1){hundreds -= 1; tens -= 12;}
else if(tens >= 2 && hundreds >= 2){hundreds -= 2; tens -= 2;}
else{loser = 0; break;}
}
loser ? puts("Hanako") : puts("Ciel");
return 0;
}Input
Output
Demonstration
Codeforces Solution-A. Bus Game-Solution in C, C++, Java, Python