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    1138 - Trailing Zeroes (III)
    Time Limit: 2 second(s) Memory Limit: 32 MB

    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'.

    Sample Input

    Output for Sample Input

    3

    1

    2

    5

    Case 1: 5

    Case 2: 10

    Case 3: impossible





    题意:给你一个数Q。代表N!中末尾连续0的个数。让你求出最小的N。



    求N!中尾连续0的个数:

    LL change(LL x){
        LL ans = 0;
        while(x){
            ans += x / 5;
            x /= 5;
        }
        return ans;
    }

    AC代码

    #include <stdio.h>
    #include <string.h>
    #include <queue>
    #include <algorithm>
    #define INF 0x3f3f3f3f
    #define LL long long
    
    LL change(LL x){
        LL ans = 0;
        while(x){
            ans += x / 5;
            x /= 5;
        }
        return ans;
    }
    
    int main (){
        int T, n;
        int k = 1;
        scanf("%d", &T);
        while(T--){
            scanf("%d", &n);
            LL l = 0, r = 100000000 * 5 + 10;
            LL mid, ans;
            while(r > l){
                mid = (l + r) / 2;
                if(change(mid) >= n){
                    ans = mid;
                    r = mid;
                }
                else
                    l = mid + 1;
            }
            printf("Case %d: ", k++);
            if(change(ans) == n)
                printf("%lld
    ", ans);
            else
                printf("impossible
    ");
    
        }
        return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/yxysuanfa/p/7110658.html
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