hdu 1316 How many Fibs?(高精度斐波那契数)

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// 大数继续

Problem Description

Recall the definition of the Fibonacci numbers:
f1 := 1
f2 := 2
fn := fn-1 + fn-2 (n >= 3)

Given two numbers a and b, calculate how many Fibonacci numbers are in the range [a, b].

Input

The input contains several test cases. Each test case consists of two non-negative integer numbers a and b. Input is terminated by a = b = 0. Otherwise, a <= b <= 10^100. The numbers a and b are given with no superfluous leading zeros.

Output

For each test case output on a single line the number of Fibonacci numbers fi with a <= fi <= b.

Sample Input

10 1001234567890 98765432100 0

Sample Output

Source

University of Ulm Local Contest 2000

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高精度斐波数，依然用高精度的加法模板打表，打到520多就够了，

主要是判断数的大小，这里是比较额字符串形式的数的大小。

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Code:

#include<iostream>#include <stdio.h>#include<string>using namespace std;string add(string x,string y){    string ans ;    int lenx = x.length();    int leny = y.length();    if(lenx<leny)    {        for(int i = 1;i<=leny-lenx;i++)            x = "0"+x;    }    else    {        for(int i = 1;i<=lenx-leny;i++)            y = "0"+y;    }    lenx = x.length();    int cf = 0;    int temp;    for(int i = lenx-1;i>=0;i--)    {        temp = x[i] - '0' + y[i] - '0'+cf;        cf = temp/10;        temp%=10;        ans = char('0'+temp)+ans;    }    if(cf!=0)        ans = char(cf+'0')+ans;    return ans;}int compare(string x,string y)//  字符串形式的数的比较大小{    int i,lenx = x.length(),leny = y.length(),leaf;    if(x==y)  return 0;//   0 表示 x == y    if(x.length()>y.length())   return 1;// 返回1 表示 x > y    if(x.length()<y.length())   return -1;// -1 表示 x < y    if(x.length()==y.length())    {        for(i = 0;i<lenx;i++)        {            if(x[i]==y[i])  continue;            if(x[i]>y[i])   return 1;            else    return -1;        }        return 0;    }    return leaf;}int main(){    int i,j,k,start,eend;    string x,y,num[1005];;    num[0] = "0";    num[1] = "1";    num[2] = "2";    for(int i = 3;i<=1000;i++)        num[i] = add(num[i-1],num[i-2]);    while(cin>>x>>y&&x!="0"||y!="0")//  x y 均为 0 的时候才结束程序    {        if(y == "0")//  y == 0  时 直接输出 0         {            printf("0");            continue;        }        start = eend = 0;        /**        j = k = 0;        while(x[j]=='0')//  受到            j++;        x = x.substr(j,x.length()-j);//  受到 hdu 1753 的影响，以为会有前导0，其实没有        while(y[k]=='0')            k++;        y = y.substr(k,y.length()-k);        **/        for(i = 1;i<1000;i++)        {            if(compare(x,num[i])==0)            {                start = i;                break;            }            else if(compare(num[i],x)==-1&&compare(num[i+1],x)==1)            {                start = i+1;                break;            }        }        for(i = 1;i<1000;i++)        {            if(compare(y,num[i])==0){                eend = i;break;            }            else if(compare(num[i],y)==-1&&compare(num[i+1],y)==1){                eend = i;break;            }        }        if(x=="0") //  注意  x == 0 时的情况            start = 1;        //cout<<start<<"      "<<eend<<endl;;        cout<<eend-start+1<<endl;    }}

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