lightoj- 1138 - Trailing Zeroes (III)【思维】【二分查找】

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1138 - Trailing Zeroes (III)

You task is to find minimal natural number N, so that N! contains exactly Q zeroes on the trail in decimal notation. As you know N! = 1*2*...*N. For example, 5! = 120, 120 contains one zero on the trail.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer Q (1 ≤ Q ≤ 10⁸) in a line.

Output

For each case, print the case number and N. If no solution is found then print 'impossible'.

题解：一个数阶乘有 n 个为 5 的因子，就会产生 n 个 0；然后就是找的这个数，刚开始各种爆啊，二分查找啊，我个ZZ；

#include<cstdio>
#include<algorithm>
#include<cstring>
#define LL long long 
using namespace std;
const int INF=0x3f3f3f3f;
int Q;
LL solve(int x)
{
  LL sum=0;
  while(x)
  {
    sum+=x/5;
    x/=5;
  }
  return sum;
}
int main()
{
  int t,text=0;
  scanf("%d",&t);
  while(t--)
  {
    scanf("%lld",&Q);
    LL mid,l=0,r=INF;
    while(l<=r)
    {
      mid=l+r>>1;
      if(solve(mid)>=Q)
        r=mid-1;
      else
        l=mid+1;
    }
    printf("Case %d: ",++text);
    if(solve(l)==Q)
      printf("%lld\n",l);
    else
      puts("impossible");
  }
  return 0;
}