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  • 大数变形版斐波那契

    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
    1**00

    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.

    #include<stdio.h>
    #include<string.h>
    int a[7500][600];
    int main()
    {
        memset(a,0,sizeof(a));
        int i,j,k,n,m;
        a[2][0]=1;
        a[1][0]=1;
        a[3][0]=1;
        a[4][0]=1;
        int g=0;
        for(i=5;i<7500;i++)
        {
            for(j=0;j<=g;j++)
            {
                a[i][j]=a[i-1][j]+a[i-2][j]+a[i-3][j]+a[i-4][j];
            }
            for(j=0;j<=g;j++)
            {
                a[i][j+1]+=a[i][j]/10000;//每位存放10000,一开始存放10,结果超内存
                a[i][j]%=10000;
            }
            if(a[i][g])
            {
                g++;
            }
        }
        int h;
        while(scanf("%d",&h)!=EOF)
        {
            for(i=g;i>=0;i--)
                if(a[h][i]!=0)
                    break;
                printf("%d",a[h][i--]);
            for(;i>=0;i--)
                printf("%04d",a[h][i]);//不足4位前置0
            printf("
    ");
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/nanfenggu/p/7900193.html
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