Codeforces 689C - Mike and Chocolate Thieves

题意：有四个小偷要偷巧克力，每个小偷偷的巧克力是前一个小偷的k倍。例：a,ak,ak^2,ak^3。然后给出你小偷头巧克力的方案数，问：符合条件的条件下，背包中巧克力量最大值最小是多少。

分析：最大值最小。所以二分n。

 1 /************************************************
 2 Author        :DarkTong
 3 Created Time  :2016/7/7 17:06:12
 4 File Name     :C.cpp
 5 *************************************************/
 6
 7 #include <cstdio>
 8 #include <cstring>
 9 #include <iostream>
10 #include <algorithm>
11 #include <vector>
12 #include <queue>
13 #include <set>
14 #include <map>
15 #include <string>
16 #include <cmath>
17 #include <cstdlib>
18 #include <ctime>
19
20 #define INF 0x3f3f3f3f
21 #define esp 1e-9
22 typedef long long ll;
23 using namespace std;
24 ll m;
25 ll Check(ll x)
26 {
27     ll sum=0;
28     for(ll i=2;;++i)
29     {
30         if(x<i*i*i) break;
31         sum+=x/(i*i*i);
32     }
33     return sum;
34 }
35 ll BS()
36 {
37     ll L=0,R=0x7fffffffffffffff;
38     ll t, mid;
39     while(L<R)
40     {
41         mid =(L+R+1)/2;
42         t = Check(mid);
43 //        cout<<"t:"<<t<<" L:"<<L<<" R:"<<R<<endl;
44         if(t==m) break;
45         else if(t>m) R=mid-1;
46         else L=mid+1;
47     }
48     if(t==m) return mid;
49     else return -1;
50 }
51 void solve(ll x)
52 {
53 //<    cout<<"x:"<<x<<endl;
54     if(x==-1) puts("-1");
55     else
56     {
57         ll ma=0;
58         for(ll i=2;;++i)
59         {
60             if(x<i*i*i) break;
61             ma = max(ma, (x/(i*i*i)*i*i*i));
62         }
63         cout<<ma<<endl;
64     }
65 }
66
67
68 int main()
69 {
70     //freopen("in.txt","r",stdin);
71     //freopen("out.txt","w",stdout);
72     scanf("%lld", &m);
73     solve(BS());
74     return 0;
75 }