Light OJ 1138-trailing Zeroes (III) "Two points to find a good question" "Give N!" At the end there is a continuous Q 0, which allows you to find the smallest n "

1138-trailing Zeroes (III)

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Time Limit: 2 second (s)

Memory Limit: MB

You task was 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! = 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 '.

Sample Input	Output for Sample Input
3 1 2 5	Case 1:5 Case 2:10 Case 3:impossible

Problem Setter:jane ALAM JAN

Test instructions: give you a number q, which means n! The number of consecutive 0 in the end. Let you find the smallest n.

Theorem: Ask N! End ofnumber of consecutive 0
Find the following method

ll sum (ll N) {    ll ans = 0;    while (N)    {        ans + = N/5;        N/= 5;    }    return ans;}

Originally dared not write, finally found even q = 10^8 will not be super long long (seemingly int is not super)

Two again, the interval is small. WA once more.


AC code: Two-point implementation

#include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <queue> #include <stack> #include <vector> #include <map> #include <string> #include <algorithm># Define LL long long#define maxn 100+10#define maxm 20000+10#define INF 0x3f3f3f3fusing namespace std;    ll sum (ll N)//For the number of consecutive 0 at the end of the N factorial {ll ans = 0;        while (N) {ans + = N/5;    N/= 5; } return ans;    int k = 1;int main () {int t;    LL Q;    scanf ("%d", &t);        while (t--) {scanf ("%lld", &q);        ll left = 1, right = 1000000000000;//begins to open small drunk ll ans = 0;            while (right >= left) {int mid = (left + right) >> 1;                if (sum (mid) = = Q)//equal to the value to be assigned to ans {ans = mid;            right = Mid-1;            } else if (sum (mid) > Q) right = mid-1;        else left = mid + 1; } printf ("Case%d:", k++);        if (ans) printf ("%lld\n", ans);    else printf ("impossible\n"); } return 0;}

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Light OJ 1138-trailing Zeroes (III) "Two points to find a good question" "Give N!" At the end there is a continuous Q 0, which allows you to find the smallest n "