ZOJ Problem Set - 1005 Jugs

文章出处：http://blog.csdn.net/rongyongfeikai2/article/details/7315702作者思路简单易懂，借鉴了~~~JugsTime Limit: 2 Seconds Memory Limit: 65536 KB Special JudgeIn the movie "Die Hard 3", Bruce Willis and Samuel L. Jackson were confronted with the following puzzle. They were given a 3-gallon jug and a 5-gallon jug and were asked to fill the 5-gallon jug with exactly 4 gallons. This problem generalizes that puzzle.You have two jugs, A and B, and an infinite supply of water. There are three types of actions that you can use: (1) you can fill a jug, (2) you can empty a jug, and (3) you can pour from one jug to the other. Pouring from one jug to the other stops when the first jug is empty or the second jug is full, whichever comes first. For example, if A has 5 gallons and B has 6 gallons and a capacity of 8, then pouring from A to B leaves B full and 3 gallons in A.A problem is given by a triple (Ca,Cb,N), where Ca and Cb are the capacities of the jugs A and B, respectively, and N is the goal. A solution is a sequence of steps that leaves exactly N gallons in jug B. The possible steps arefill A fill B empty A empty B pour A B pour B A successwhere "pour A B" means "pour the contents of jug A into jug B", and "success" means that the goal has been accomplished.You may assume that the input you are given does have a solution.InputInput to your program consists of a series of input lines each defining one puzzle. Input for each puzzle is a single line of three positive integers: Ca, Cb, and N. Ca and Cb are the capacities of jugs A and B, and N is the goal. You can assume 0 < Ca <= Cb and N <= Cb <=1000 and that A and B are relatively prime to one another.OutputOutput from your program will consist of a series of instructions from the list of the potential output lines which will result in either of the jugs containing exactly N gallons of water. The last line of output for each puzzle should be the line "success". Output lines start in column 1 and there should be no empty lines nor any trailing spaces.

Sample Input

3 5 4
5 7 3

Sample Output

fill B
pour B A
empty A
pour B A
fill B
pour B A
success
fill A
pour A B
fill A
pour A B
empty B
pour A B
success

思路：由于bottle2的容量一定大于bottle1的容量，所以bottle2一定可以最终放入指定的水量。要解这道题目，按照以下算法：1.判断bottle1中是否有水，没有水就灌满水2.将bottle1中的水放入bottle2中3.判断bottle2是否等于指定水量，如果没有，那么就将bottle2清空

#include <stdio.h>int main(){	int a, b, p;	int bottle1, bottle2;	while (scanf("%d%d%d", &a, &b, &p) != EOF){		bottle1 = 0;		bottle2 = 0;		while (1){			if (bottle1 == 0){				bottle1 = a;				printf("fill A\n");			}			else if (bottle1 + bottle2 <= b){				bottle2 += bottle1;				bottle1 = 0;				printf("pour A B\n");			}			else{				bottle1 = bottle1 - (b - bottle2);				bottle2 = b;				printf("pour A B\n");			}						if (bottle2 == p){				printf("success\n");				break;			}			if (bottle2 == b){				bottle2 = 0;				printf("empty B\n");			}		}	}	return 0;}