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  • hdu 3054 Fibonacci 找循环节的公式题

    Fibonacci

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)


    Problem Description
    We know the Fibonacci Sequence

    F1=1,F2=1,F3=2,F4=3,F5=5,

    ...

    Fx = Fx-1+Fx-2

    We want to know the Mth number which has K consecutive "0" at the end of Fx.

    For example,

    F15=610

    It is the first number which has only one "0" at the end.

    F300=222232244629420445529739893461909967206666939096499764990979600.

    It is the second number which has two "0" at the end.

    Of course, the Fx may be very large if M and K are big. So we only want to know the subscript of Fx (it means the "x" For a given M and K)
     
    Input
    Input includes multiple cases.

    First line is the number of case x

    The next x lines: Each line contains two integer number, K and M, divided by a space.
     
    Output
    For each case:

    Print a integer number in a line, is the Mth number which has K consecutive 0s at the end of Fx. (You can believe the answer is smaller than 2^31);
     
    Sample Input
    3 1 1 2 2 2 5
     
    Sample Output
    15 300 900
     
    Source
    #include<bits/stdc++.h>
    using namespace std;
    #define ll long long
    #define pi (4*atan(1.0))
    #define eps 1e-14
    const int N=2e5+10,M=4e6+10,inf=1e9+10,MOD=1000;
    const ll INF=1e18+10;
    int a[100]={1,15,150,750,7500,75000,750000,7500000,75000000,750000000};
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
            int k,m;
            scanf("%d%d",&k,&m);
            if(k==2)
                printf("%d
    ",((m-1)/4*5+1)*a[k]+((m-1)%4)*a[k]);
            else
                printf("%d
    ",((m-1)/9*10+1)*a[k]+((m-1)%9)*a[k]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jhz033/p/6005689.html
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