ZOJ-1005* 灌水问题
1005：给定两个不同容量的杯子A,B和目标水量，求倒水序列。设A<B
思路：一种简便方法,只要A为空就倒满A,将A倒向B,B溢出则倒空。直到符合要求。
#include<stdio.h>#include<string.h>#include<iostream>using namespace std;int main(){int A;int B;int target;int inA;int inB;while(scanf("%d%d%d",&A,&B,&target)!=EOF){inA=0;inB=0;while(1){if(inA==0){printf("fill A\n");inA=A;}else{printf("pour A B\n");inB+=inA;if(inB>=B){if(B!=target){printf("empty B\n");inA=inB-B;inB=0;}elseinB=B;}else{inA=0;}}if(inA==target||inB==target){printf("success\n");break;}}}}