UVA 10183 How Many Fibs?

高精度推出大概600项fabi数，就包含了题目的数据范围，对于每组a,b，从1到600枚举res[i]即可

可以直接JAVA大数。我自己时套了C++高精度的版

JAVA 复制别人的

import java.math.BigInteger;

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {

        Scanner cin = new Scanner(System.in);

        BigInteger[] f = new BigInteger[600];

        f[0] = new BigInteger("1");

        f[1] = new BigInteger("2");

        for(int i = 2; i < 600; i ++)

            f[i] = f[i - 1].add(f[i - 2]);

        for(;;)

        {

            BigInteger a, b;

            int res = 0;

            a = cin.nextBigInteger();

            b = cin.nextBigInteger();

            if(a.compareTo(BigInteger.ZERO) == 0 && b.compareTo(BigInteger.ZERO) == 0)

                break;

            for(int i = 0; i < 600; i ++)

                if(f[i].compareTo(a) != -1 && f[i].compareTo(b) != 1)

                    res ++;

            System.out.println(res);

        }

    }

}

C++

#include <map>

#include <set>

#include <list>

#include <cmath>

#include <ctime>

#include <deque>

#include <stack>

#include <queue>

#include <cctype>

#include <cstdio>

#include <string>

#include <vector>

#include <climits>

#include <cstdlib>

#include <cstring>

#include <iostream>

#include <algorithm>

#define LL long long

#define PI 3.1415926535897932626

using namespace std;

int gcd(int a, int b) {return a % b ==  ? b : gcd(b, a % b);}

const int numlen=;

struct bign {

    int len, s[numlen];

    bign() {

        memset(s, , sizeof(s));

        len = ;

    }

    bign(int num) { *this = num; }

    bign(const char *num) { *this = num; }

    bign operator = (const int num) {

        char s[numlen];

        sprintf(s, "%d", num);

        *this = s;

        return *this;

    }

    bign operator = (const char *num) {

        len = strlen(num);

        while(len >  && num[] == '') num++, len--;

        for(int i = ;i < len; i++) s[i] = num[len-i-] - '';

        return *this;

    }

    void deal() {

        while(len >  && !s[len-]) len--;

    }

    bign operator + (const bign &a) const {

        bign ret;

        ret.len = ;

        int top = max(len, a.len) , add = ;

        for(int i = ;add || i < top; i++) {

            int now = add;

            if(i < len) now += s[i];

            if(i < a.len)   now += a.s[i];

            ret.s[ret.len++] = now%;

            add = now/;

        }

        return ret;

    }

    bign operator - (const bign &a) const {

        bign ret;

        ret.len = ;

        int cal = ;

        for(int i = ;i < len; i++) {

            int now = s[i] - cal;

            if(i < a.len)   now -= a.s[i];

            if(now >= )    cal = ;

            else {

                cal = ; now += ;

            }

            ret.s[ret.len++] = now;

        }

        ret.deal();

        return ret;

    }

    bign operator * (const bign &a) const {

        bign ret;

        ret.len = len + a.len;

        for(int i = ;i < len; i++) {

            for(int j = ;j < a.len; j++)

                ret.s[i+j] += s[i]*a.s[j];

        }

        for(int i = ;i < ret.len; i++) {

            ret.s[i+] += ret.s[i]/;

            ret.s[i] %= ;

        }

        ret.deal();

        return ret;

    }

    bign operator * (const int num) {

//        printf("num = %d\n", num);

        bign ret;

        ret.len = ;

        int bb = ;

        for(int i = ;i < len; i++) {

            int now = bb + s[i]*num;

            ret.s[ret.len++] = now%;

            bb = now/;

        }

        while(bb) {

            ret.s[ret.len++] = bb % ;

            bb /= ;

        }

        ret.deal();

        return ret;

    }

    bign operator / (const bign &a) const {

        bign ret, cur = ;

        ret.len = len;

        for(int i = len-;i >= ; i--) {

            cur = cur*;

            cur.s[] = s[i];

            while(cur >= a) {

                cur -= a;

                ret.s[i]++;

            }

        }

        ret.deal();

        return ret;

    }

    bign operator % (const bign &a) const {

        bign b = *this / a;

        return *this - b*a;

    }

    bign operator += (const bign &a) { *this = *this + a; return *this; }

    bign operator -= (const bign &a) { *this = *this - a; return *this; }

    bign operator *= (const bign &a) { *this = *this * a; return *this; }

    bign operator /= (const bign &a) { *this = *this / a; return *this; }

    bign operator %= (const bign &a) { *this = *this % a; return *this; }

    bool operator < (const bign &a) const {

        if(len != a.len)    return len < a.len;

        for(int i = len-;i >= ; i--) if(s[i] != a.s[i])

            return s[i] < a.s[i];

        return false;

    }

    bool operator > (const bign &a) const  { return a < *this; }

    bool operator <= (const bign &a) const { return !(*this > a); }

    bool operator >= (const bign &a) const { return !(*this < a); }

    bool operator == (const bign &a) const { return !(*this > a || *this < a); }

    bool operator != (const bign &a) const { return *this > a || *this < a; }

    string str() const {

        string ret = "";

        for(int i = ;i < len; i++) ret = char(s[i] + '') + ret;

        return ret;

    }

};

istream& operator >> (istream &in, bign &x) {

    string s;

    in >> s;

    x = s.c_str();

    return in;

}

ostream& operator << (ostream &out, const bign &x) {

    out << x.str();

    return out;

}

bign res[];

bign a,b;

void init()

{

    res[]="";

    res[]="";

    for (int i=;i<;i++)

        res[i]=res[i-]+res[i-];

}

int main()

{

    init();

    while (cin>>a>>b)

    {

        if (a=="" && b=="") break;

        int ans=;

        for (int i=;i<;i++)

            if (res[i]>=a && res[i]<=b) ans++;

        printf("%d\n",ans);

    }

    return ;

}