} } 3.
}
if(result == 110)
System.out.println(string)
import java.util.Scanner;
public class C33 {
public static void main(String[] args){ }
static int[] getNum() { }
static void fenjiu(int num1,int num2,int num3,int num[]) {
Scanner readerScanner = new Scanner(System.in);
String string[] = readerScanner.nextLine().split(\); int i = 0;
int num[] = new int[7]; while(i < 7) { }
return num;
num[i] = Integer.parseInt(string[i]); i++;
int num[] = getNum();
int d = gcd(num[1], num[2]);//中瓶和小瓶容量的最大公约数 if(num[0]/2 <= num[1] + num[2] && (num[0]/2)%d == 0) { } else { }
System.out.println(num[3] + \ + num[4] + \ + num[5]); fenjiu(num[3], num[4], num[5],num);
System.out.println(\不可能\);
// // // // //
//C2*X - C3*Y = 1
int b1 = num[1],c1 = num[2];
int a = num1,b = num2,c = num3,result = num[6]; if(b == result||c == result||a == result) System.out.print(\); else {
//12,8,5,12,0,0,6 1. 大瓶子只能倒入中瓶子 2. 中瓶子只能倒入小瓶子 3. 小瓶子只能倒入大瓶子
4. 小瓶子只有在已经装满的情况下才能倒入大瓶
5. 若小瓶子被倒空，则无论中瓶子是否满，应马上从中瓶子倒入小瓶子
if(c == c1) {//
4. 小瓶子只有在已经装满的情况下才能倒入大瓶子
a = a + c; c = 0;
System.out.println(a + \ + b + \ + c); fenjiu(a, b, c, num);
} else { if(c == 0 && b > 0) {// 5. 若小瓶子被倒空，则无论中瓶子是否满，应马上从中瓶子倒入小瓶子 if(b > c1) {
c = c1;
b = b - c1;
System.out.println(a + \ + b + \ + c); fenjiu(a, b, c, num); } else { c = b; b = 0;
System.out.println(a + \ + b + \ + c);
}
}
}
} }
}
fenjiu(a, b, c, num);
else { }
if(b != b1) { } else { }
b = b - c1 + c; c = c1;
System.out.println(a + \ + b + \ + c); fenjiu(a, b, c, num); a = a - b1 + b; b = b1;
System.out.println(a + \ + b + \ + c); fenjiu(a, b, c, num);
public static int gcd(int m, int n) { }
int a = n, b = m; { }
if ((a = a % b) == 0)
return b; return a;
if ((b = b % a) == 0)
while (true)
本答案仅供参考，如有不同意见属正常。希望本答案对您的学习有帮助！谢谢参阅！