BZOJ 4029 [HEOI 4029] 定价 解题报告

这个题好像也是贪心的感觉。。

我们枚举 $1,5,10,50,100,\dots$ ，找出在 $[l, r]$ 内能整除它们的最小的数。

然后找到其中在荒谬值最小的情况下数值最小的那个数，

就做完了。

 #include <cmath>
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;
 typedef long long LL;
 #define INF 593119681

 int _;
 LL l, r;

 inline LL Calc(LL x)
 {
     if (!~x) return INF;
     while (x %  == ) x /= ;
     int f = (x % ) ?  : -;
     int res = ;
     for (; x; x /= )
         res += ;
     return res + f;
 }

 int main()
 {
     #ifndef ONLINE_JUDGE
         freopen("4029.in", "r", stdin);
         freopen("4029.out", "w", stdout);
     #endif

     for (scanf("%d", &_); _; _ --)
     {
         scanf("%lld%lld", &l, &r);
         LL best = -, top, bot;
         for (LL x = , _x = ; x <= r; x *= , _x *= )
         {
             top = (l + x - ) / x * x;
             bot = r / x * x;
             if (top <= bot) best = (Calc(best) < Calc(top) || (Calc(best) == Calc(top) && best < top)) ? best : top;
             top = (l + _x - ) / _x * _x;
             bot = r / _x * _x;
             if (top <= bot) best = (Calc(best) < Calc(top) || (Calc(best) == Calc(top) && best < top)) ? best : top;
         }
         printf("%lld\n", best);
     }

     #ifndef ONLINE_JUDGE
         fclose(stdin);
         fclose(stdout);
     #endif
     return ;
 }

4029_Gromah