1103

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

has 4 digits.

来源

19 June, 2011 - Waterloo local contest

#include<stdio.h>typedef struct fanum{char fnum[1001];int len;}Fanum;void strAdd(Fanum *sf1,Fanum *sf2){int i,len,flag=0;Fanum sf3;len=sf1->len>sf2->len?sf1->len:sf2->len;sf3.len=len;for(i=0;i<len;i++){sf3.fnum[i]= sf1->fnum[i]-'0'+sf2->fnum[i]-'0'+flag;flag=sf3.fnum[i]/10;sf3.fnum[i]= sf3.fnum[i]%10+'0';//转换两次}if(flag==1){sf3.fnum[i]='1';sf3.len++;}for(i=0;i<sf3.len;i++){sf1->fnum[i]=sf2->fnum[i];sf1->len=sf2->len;sf2->fnum[i]=sf3.fnum[i];sf2->len=sf3.len;}return;}int main(){Fanum f1,f2;long num,i;scanf("%d",&num);f1.len= 1;f2.len= 1;for(i=0;i<1001;i++){f1.fnum[i]='0';f2.fnum[i]='0';}f1.fnum[0]='1';f2.fnum[0]='1';while(num!=2){strAdd(&f1,&f2);num-- ;}for(i=f2.len-1;i>=0;i--)printf("%d",f2.fnum[i]-'0');printf("\n");return 1;}