Hat's Fibonacci

题目链接
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
100
Sample Output
4203968145672990846840663646
题意:斐波那契数列,大数模拟
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>
#define N 10000 
int str[N][260];
int main()
{
    memset(str,0,sizeof(str));
    str[1][0]=1;
    str[2][0]=1;
    str[3][0]=1;
    str[4][0]=1;
    int i,j,ans=0,c,n;
    for(i=5;i<N;i++)
    {
        for(j=0,c=0;j<260;j++)
        {
            ans=str[i-1][j]+str[i-2][j]+str[i-3][j]+str[i-4][j]+c;
            c=ans/100000000;
            str[i][j]=ans%100000000;  // 每一个数组存8位数字,c来控制是否进位。
        }
    }
    while(scanf("%d",&n)!=EOF)
    {
        j=259;
        while(!str[n][j])    //将首0清除。
        j--;
        printf("%d",str[n][j]);
        for(i=j-1;i>=0;i--)
        printf("%08d",str[n][i]);//每8位输出,不足8位自动补0;
        printf("\n");
    }
    return 0;
}
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11-08 16:53
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滑模小马达:第三个如果是qfqc感觉还行,我签的qfkj搞电机的,违约金也很高,但公司感觉还可以,听说之前开过一个试用转正的应届生,仅供参考。
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