PDSOJ 1048 Fibonacci Numbers(大数)

Fibonacci Numbers

时间限制: 1 Sec  内存限制: 128 MB
提交: 13  解决: 2
[提交][状态][讨论版]

题目描述

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.

输入

100

输出

354224848179261915075

样例输入

100

样例输出

354224848179261915075

提示

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

和大数加法类似，还要注意数据的变换s1,s2,sum之间的变换，sum=s1+s2.然后字符处理

#include<iostream>#include<cstring>using namespace std;char sum[1200];int s=0,m=0,n;int main(){    cin>>n;    string s1,s2;    int a[10000],b[10000];    int he,i;    s1="1";    s2="1";    for(m=2;m<n;m++)    {        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        a[0]=s1.length();        for(i=1;i<=a[0];i++)        {            a[i]=s1[a[0]-i]-'0';        }          b[0]=s2.length();        for(i=1;i<=b[0];i++)        {            b[i]=s2[b[0]-i]-'0';        }        he=(a[0]>b[0]?a[0]:b[0]);        for(i=1;i<=he;i++)        {            a[i]=a[i]+b[i];            a[i+1]=a[i+1]+a[i]/10;            a[i]=a[i]%10;        }        he++;        while((a[he]==0)&&(he>1))        he--;        for(i=he,s=0;i>=1;i--,s++)        {            sum[s]=a[i]+'0';        }        s1=s2;        s2=sum;   }        cout<<s2<<endl;}