HDU1250大数+斐波那契数列

Posted 2020-06-14

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

100

 

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.

 

这题并非看起来那么简单（如果你系统的学习C而惯用C++）

原因至今还没找到，如果明白以后还会继续补充这道题。

我用cout代替printf（用的是别人正确的代码）也会WA。

再也不敢乱用C++了

#include<iostream>
#include<stdio.h>
#include<cstring>
#define N 10000
#define M 300
using namespace std;

int f[N][M];

void work()
{
    memset(f,0,sizeof(f));
    f[1][1]=f[2][1]=f[3][1]=f[4][1]=1;
    int i,j,t;
    for(i=5;i<N;i++)
    {
        t=0;
        for(j=1;j<M;j++)
        {
            t=t+f[i-1][j]+f[i-2][j]+f[i-3][j]+f[i-4][j];
            f[i][j]=t%100000000;
            t=t/100000000;
        }
    }
}

int main()
{
    int n,i;
    work();
    while(scanf("%d",&n)!=EOF)
    {
        i=M-1;
        while(f[n][i]==0)i--;
        printf("%d",f[n][i--]);
        while(i>=1)
        {
            printf("%08d",f[n][i]);
            i--;
        }
        printf("\n");
    }
    return 0;
}