UVa 10465 Homer Simpson: Enumeration

10465-homer Simpson

Time limit:3.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=114&page=show_ problem&problem=1406

This amount of data. Enumerations are definitely the quickest way to go.

Complete code:

/*0.025s*/
    
#include <bits/stdc++.h>  
using namespace std;  
    
int main ()  
{  
    int m, n, T, X, ANS1, ans2, drink, I, K;  
    while (~SCANF ("%d%d%d", &m, &n, &t))  
    {  
        if (M > N) swap (m, n);  
        x = t/m, drink = -1u>>1;  
        for (i = x; I >= 0 && drink-i)///enum  
        {  
            k = t-i * m;  
            if (Drink > K% n) drink = k% n, ans1 = i + k/n;///drinks as little beer as possible  
            if (k% n = 0) Ans2 = i + k/n;  
        }  
        if (drink) printf ("%d%d\n", ans1, drink);  
        else printf ("%d\n", ans2);  
    }  
    return 0;  
}

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