BZOJ2490 Zombie’s Treasure Chest

如果n = lcm(s1, s2)，那么就可以直接得到maxV = (v / s1 * v1, v / s2 *v2)

然后还剩下一点体积我们暴力枚举用s1的量，让s1为max(s1, s2)可以减少枚举次数。。。然后就做完了QAQ

 /**************************************************************
     Problem: 2490
     User: rausen
     Language: C++
     Result: Accepted
     Time:4 ms
     Memory:804 kb
 ****************************************************************/

 #include <cstdio>
 #include <algorithm>

 using namespace std;
 typedef long long ll;

 ll v, s1, s2, v1, v2;
 ll ans, del;

 int main() {
     int T, icase, i;
     ll t, tmp;
     scanf("%d", &T);
     for (icase = ; icase <= T; ++icase) {
         scanf("%lld%lld%lld%lld%lld", &v, &s1, &v1, &s2, &v2);
         if (s1 < s2) swap(s1, s2), swap(v1, v2);
         tmp = s1 * s2 / __gcd(s1, s2), t = v / tmp;
         v = v % tmp;
         if (t) --t, v += tmp;
         ans = max(t * (tmp / s1) * v1, t * (tmp / s2) * v2);
         for (i = del = ; i <= v / s1; ++i)
             del = max(del, i * v1 + (v - i * s1) / s2 * v2);
         printf("Case #%d: %lld\n", icase, ans + del);
     }
     return ;
 }