Fibonacci NumbersTime

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题目地址：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1828

Time Limit: 2 Seconds      Memory Limit: 65536 KB

A Fibonacci sequence is calculated by adding the previous two members of the sequence, with the first two members being both 1.

f(1) = 1, f(2) = 1, f(n > 2) = f(n - 1) + f(n - 2)

Your task is to take a number as input, and print that Fibonacci number.

Sample Input

100

Sample Output

354224848179261915075

Note:

No generated Fibonacci number in excess of 1000 digits will be in the test data, i.e. f(20) = 6765 has 4 digits.

考察高精度运算

关键点是求每一位相加后的数字，并且还应该保证在输出的时候尽可能的方便。

这个时候关键点就是对a[i][j]进行对10取余，这样就保证了数组正序存放的数是逆序的数据。

这道题竟然是多组输入，找了 很久BUG，发现是输出错误~~~~~~

#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
#include<cstdio>
using namespace std;
int a[5010][1010];
void init()
{
    a[1][0]=1;
    a[2][0]=1;
    for(int i=3; i<=5001; ++i)
    {
       int c=0;
       for(int j=0;j<=1001;++j)
       {
           a[i][j]=(a[i-1][j]+a[i-2][j]+c)%10;
           c=(a[i-1][j]+a[i-2][j]+c)/10;
       }
    }
}
int main()
{
    init();
    int n;
    while(~scanf("%d",&n))
    {
    int i;
    for(i=1001;i>=0;--i)
        if(a[n][i]!=0)
            break;
    for(;i>=0;--i)
    {
        printf("%d",a[n][i]);
    }
    printf("\n");
    }
    return 0;
}