hdu3054（斐波那契。。。。找规律）

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

31 12 22 5 

 

Sample Output

15300900 

打表。。。。这题代码量比较少，但还是很考验找规律能力的，而且！！！
眼力必须好啊！

下面是我的打表代码，输入n代表末尾n个0：

#include <iostream>using namespace std;int pp(int n){    int ans=1;    for(int i=0;i<n;i++)        ans*=10;    return ans;}int f[100000000]={0};int main(){    int n;    while(cin>>n)    {        int p0=pp(n),p1=pp(n+1),t=1;        f[1]=1;        for(int i=2;i<100000000;i++)        {            f[i]=(f[i-1]+f[i-2])%p1;            if(f[i]%p0==0&&f[i]!=0)            {                cout<<i<<"th feibo       "<<t++<<"th number"<<endl;            }            if(t==50)break;        }    }    return 0;}

运行完打表代码之后会发现1,3,4,5,6,7，……都是到第9个数增量是有一个变化！

而2是到第4个数增量有了变化！（说实话真的很难仔细一直看到第9个！！能看出2都很不错了）

然后就是找规律了，这个自己都能写的！

#include <iostream>#include <stdio.h>#include<string.h>using namespace std;int pp(int n){    int ans=1;    for(int i=0;i<n;i++)        ans*=10;    return ans;}int main(){    int t;   scanf("%d",&t);        while(t--)    {           int k,m;           scanf("%d%d",&k,&m);        if(k==1)        {            printf("%d\n",15*(((m-1)/9)*10+1)+15*((m-1)%9));        }        else if(k==2)        {            printf("%d\n",150*((m-1)/4*5+1)+150*((m-1)%4));        }        else        {            printf("%d\n",75*pp(k-2)*(((m-1)/9)*10+1)+75*pp(k-2)*((m-1)%9));        }    }    return 0;}