Problem C: | Homer Simpson |
Time Limit: 3 seconds Memory Limit: 32 MB |
Homer Simpson, a very smart guy, likes eating Krusty-burgers. It takes Homer m minutes to eat a Krusty- burger. However, there�s a new type of burger in Apu�s Kwik-e-Mart. Homer likes those too. It takes him n minutes to eat one of these burgers. Given t minutes, you have to find out the maximum number of burgers Homer can eat without wasting any time. If he must waste time, he can have beer. |
Input consists of several test cases. Each test case consists of three integers m, n, t (0 < m,n,t < 10000). Input is terminated by EOF.
For each test case, print in a single line the maximum number of burgers Homer can eat without having beer. If homer must have beer, then also print the time he gets for drinking, separated by a single space. It is preferable that Homer drinks as little beer as possible.
3 5 54 3 5 55
18 17
Problem setter: Sadrul Habib Chowdhury
Solution author: Monirul Hasan (Tomal)
Time goes, you say? Ah no!
Alas, Time stays, we go.
-- Austin Dobson
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啊好水好水
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#include <iostream> #include <cstdio> #include <cstring> using namespace std; int n,m,t; int f[11111]; int main() { while (~scanf("%d%d%d",&n,&m,&t)) { memset(f,-1,sizeof(f)); f[0]=0; for (int i=0;i<=t-n;i++) { if (f[i]>=0) { f[i+n]=max( f[i+n],f[i]+1 ); } } for (int i=0;i<=t-m;i++) { if (f[i]>=0) { f[i+m]=max( f[i+m],f[i]+1 ); } } if (f[t]!=-1) { printf("%d\n",f[t]); } else { int ans=0; int tmp=0; for (int i=t;i>=0;i--) { if (f[i]!=-1) { ans=f[i]; tmp=i; break; } } printf("%d %d\n",ans,t-tmp); } } return 0; }