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。
1 #include <cmath> 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 #include <algorithm> 6 using namespace std; 7 typedef long long LL; 8 #define INF 593119681 9 10 int _; 11 LL l, r; 12 13 inline LL Calc(LL x) 14 { 15 if (!~x) return INF; 16 while (x % 10 == 0) x /= 10; 17 int f = (x % 5) ? 0 : -1; 18 int res = 0; 19 for (; x; x /= 10) 20 res += 2; 21 return res + f; 22 } 23 24 int main() 25 { 26 #ifndef ONLINE_JUDGE 27 freopen("4029.in", "r", stdin); 28 freopen("4029.out", "w", stdout); 29 #endif 30 31 for (scanf("%d", &_); _; _ --) 32 { 33 scanf("%lld%lld", &l, &r); 34 LL best = -1, top, bot; 35 for (LL x = 1, _x = 5; x <= r; x *= 10, _x *= 10) 36 { 37 top = (l + x - 1) / x * x; 38 bot = r / x * x; 39 if (top <= bot) best = (Calc(best) < Calc(top) || (Calc(best) == Calc(top) && best < top)) ? best : top; 40 top = (l + _x - 1) / _x * _x; 41 bot = r / _x * _x; 42 if (top <= bot) best = (Calc(best) < Calc(top) || (Calc(best) == Calc(top) && best < top)) ? best : top; 43 } 44 printf("%lld ", best); 45 } 46 47 #ifndef ONLINE_JUDGE 48 fclose(stdin); 49 fclose(stdout); 50 #endif 51 return 0; 52 }