UVa 571 jugs (Idea question)

571-jugs

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem &problem=512

Ideas:

Since it is special judge, it is possible to construct a feasible solution.

Conclusion: If A is empty, add water, not empty to B pour, b full after the empty off, so in B can certainly form a 0~b any solution.

Proof: In this way to pour water, we can use (k*a)%b to indicate the amount of water in B, because of A and B coprime, so the LCM (a,b) =ab, then A accumulation of B can return to the starting point, so the minimum period is B. And, in a cycle, we can use contradiction to prove that all the values are different (as long as the last value and the previous one is not available), so we can certainly get a sequence of operations that satisfies the requirements.

Complete code:

/*0.029s*/
    
#include <cstdio>  
    
int main ()  
{  
    int na, NB, N, CA, CB;  
    while (~SCANF ("%d%d%d", &ca, &CB, &n))  
    {  
        na = 0, nb = 0;  
        while (NB!= N)  
        {  
            if (Na = = 0)  
            {  
                puts ("Fill A");  
                NA = CA;  
            }  
            if (NB = = cb)  
            {  
                puts ("Empty B");  
                NB = 0;  
            }  
            NB = na;  
            Puts ("Pour A B");  
            if (NB > cb) na = NB-CB, NB = cb;  
            else na = 0;  
        }  
        Puts ("Success");  
    }  
    return 0;  
}

Author: csdn Blog Synapse7

See more highlights of this column: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/