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
1 /************************************************************** 2 Problem: 2490 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:4 ms 7 Memory:804 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 13 using namespace std; 14 typedef long long ll; 15 16 ll v, s1, s2, v1, v2; 17 ll ans, del; 18 19 int main() { 20 int T, icase, i; 21 ll t, tmp; 22 scanf("%d", &T); 23 for (icase = 1; icase <= T; ++icase) { 24 scanf("%lld%lld%lld%lld%lld", &v, &s1, &v1, &s2, &v2); 25 if (s1 < s2) swap(s1, s2), swap(v1, v2); 26 tmp = s1 * s2 / __gcd(s1, s2), t = v / tmp; 27 v = v % tmp; 28 if (t) --t, v += tmp; 29 ans = max(t * (tmp / s1) * v1, t * (tmp / s2) * v2); 30 for (i = del = 0; i <= v / s1; ++i) 31 del = max(del, i * v1 + (v - i * s1) / s2 * v2); 32 printf("Case #%d: %lld ", icase, ans + del); 33 } 34 return 0; 35 }