ACM--steps--dyx--Hat's Fibonacci-优快云博客

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Hat's Fibonacci

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1657 Accepted Submission(s): 647

Problem Description

A Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1.
F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)
Your task is to take a number as input, and print that Fibonacci number.

Input

Each line will contain an integers. Process to end of file.

Output

For each case, output the result in a line.

Sample Input

Sample Output

4203968145672990846840663646


Note:
No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.

Author

戴帽子的

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{
    memset(dyx,0,sizeof(dyx));
    int t;
    dyx[1][1]=dyx[2][1]=dyx[3][1]=dyx[4][1]=1;
    for(int i=5;i<N;i++)
    {
        t=0;
        for(int j=1;j<M;j++)
        {
            t+=dyx[i-1][j]+dyx[i-2][j]+dyx[i-3][j]+dyx[i-4][j];
            dyx[i][j]=t%100000000;
            t/=100000000;//进位
        }
    }
}
int main()
{
    int num;
    wyx();
    while(cin>>num)
    {
        int i=M-1;
        while(dyx[num][i]==0)
        i--;
        cout<<dyx[num][i--];
        while(i>0)
        {
            printf("%08d",dyx[num][i]);
            //cout<<setw(8)<<dyx[num][i--];
            i--;
        }
        cout<<endl;
    }
    return 0;
}