hdu1316(大数的斐波那契数)

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题目信息:求两个大数之间的斐波那契数的个数(C++/JAVA)

http://acm.hdu.edu.cn/showproblem.php?pid=1316

这里给出java代码和c++代码

C++：AC代码

#include<iostream>
#include<string>
using namespace std;
string add(string s1,string s2){//字符串模拟大数加法
    string s;
    int len1,len2;
    len1=s1.size()-1; len2=s2.size()-1;
    int i=0,flag=0;
    while(len1>-1&&len2>-1){
        int sum=flag+(s1[len1--]-'0')+(s2[len2--]-'0');
        s+=char ((sum)%10+'0');
        flag=sum/10;
    }
    while(len1>-1){
        int sum=flag+(s1[len1--]-'0');
        s+=char ((sum)%10+'0');
        flag=sum/10;
    }
    while(len2>-1){
        int sum=flag+(s2[len2--]-'0');
        s+=char ((sum)%10+'0');
        flag=sum/10;
    }
    if(flag) s+=char ('0'+flag);
    //cout<<s<<endl;
    for(int i=0;i<s.size()/2;i++){
        char c=s[i];
        s[i]=s[s.size()-i-1];
        s[s.size()-i-1]=c;
    }
    return s;
}
int compareto(string s,string s1){//模拟大数(字符串)比较

    int j,i,len=s.size(),len1=s1.size();
    for(i=0;s[i]=='0';i++);//去掉前导0
    for(j=0;s1[j]=='0';j++);//去掉前导0
    if(len-i>len1-j) return 1;
    else if(len-i<len1-j) return -1;
    else for(;s[i]==s1[j]&&i<len;i++,j++);
    if(s[i]>s1[j]) return 1;//s>s1
    if(s[i]<s1[j]) return -1;//s<s1
    return 0;
}
string s[501];
int main()
{
    string a,b;
    s[1]="1"; s[2]="2";
    for(int i=3;i<=500;i++){
        s[i]=add(s[i-1],s[i-2]);
        //if(i<10) cout<<s[i]<<endl;
    }
    while(cin>>a>>b){
        if(a=="0"&&b=="0") break;
        //cout<<compareto(a,b)<<endl;
        int sum=0;
        for(int i=1;i<=500;i++){
            if(compareto(s[i],a)>=0&&compareto(s[i],b)<=0){
                sum++;
            }
        }
        cout<<sum<<endl;
    }

    return 0;
}

java代码：

import java.math.BigInteger;
import java.util.Scanner;

public class Main {

   public static void main(String[] args) {
       BigInteger a,b;
       BigInteger s[] = new BigInteger[1501];
       s[0]=BigInteger.ZERO;
       s[1]=BigInteger.ONE;
       s[2]=BigInteger.valueOf(2);
       for(int i=3;i<=1500;i++){
           s[i]=s[i-1].add(s[i-2]);
       }
       //System.out.println(s[1500]);
       Scanner sc = new Scanner(System.in);
       while(sc.hasNext()){
           a=sc.nextBigInteger();
           b=sc.nextBigInteger();
           if(a.add(b).compareTo(BigInteger.ZERO)==0) break;
           int cnt=0;
           //提高程序的鲁棒性
           if(a.compareTo(b)>0){
               BigInteger tmp=a;
               a=b;
               b=tmp;
           }
           for(int i=1;i<=1500;i++){
               if(s[i].compareTo(a)>=0&&s[i].compareTo(b)<=0){
                   cnt++;
               }
           }
           System.out.println(cnt);
       }

   }
}

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