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 == 0 ? b : gcd(b, a % b);}
const int numlen=110;
struct bign {
    int len, s[numlen];
    bign() {
        memset(s, 0, sizeof(s));
        len = 1;
    }
    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 > 1 && num[0] == ‘0‘) num++, len--;
        for(int i = 0;i < len; i++) s[i] = num[len-i-1] - ‘0‘;
        return *this;
    }

    void deal() {
        while(len > 1 && !s[len-1]) len--;
    }

    bign operator + (const bign &a) const {
        bign ret;
        ret.len = 0;
        int top = max(len, a.len) , add = 0;
        for(int i = 0;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%10;
            add = now/10;
        }
        return ret;
    }
    bign operator - (const bign &a) const {
        bign ret;
        ret.len = 0;
        int cal = 0;
        for(int i = 0;i < len; i++) {
            int now = s[i] - cal;
            if(i < a.len)   now -= a.s[i];
            if(now >= 0)    cal = 0;
            else {
                cal = 1; now += 10;
            }
            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 = 0;i < len; i++) {
            for(int j = 0;j < a.len; j++)
                ret.s[i+j] += s[i]*a.s[j];
        }
        for(int i = 0;i < ret.len; i++) {
            ret.s[i+1] += ret.s[i]/10;
            ret.s[i] %= 10;
        }
        ret.deal();
        return ret;
    }

    bign operator * (const int num) {
//        printf("num = %d\n", num);
        bign ret;
        ret.len = 0;
        int bb = 0;
        for(int i = 0;i < len; i++) {
            int now = bb + s[i]*num;
            ret.s[ret.len++] = now%10;
            bb = now/10;
        }
        while(bb) {
            ret.s[ret.len++] = bb % 10;
            bb /= 10;
        }
        ret.deal();
        return ret;
    }

    bign operator / (const bign &a) const {
        bign ret, cur = 0;
        ret.len = len;
        for(int i = len-1;i >= 0; i--) {
            cur = cur*10;
            cur.s[0] = 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-1;i >= 0; 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 = 0;i < len; i++) ret = char(s[i] + ‘0‘) + 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[600];
bign a,b;
void init()
{
    res[1]="1";
    res[2]="2";
    for (int i=3;i<600;i++)
        res[i]=res[i-1]+res[i-2];
}
int main()
{
    init();
    while (cin>>a>>b)
    {
        if (a=="0" && b=="0") break;
        int ans=0;
        for (int i=1;i<600;i++)
            if (res[i]>=a && res[i]<=b) ans++;
        printf("%d\n",ans);
    }
    return 0;
}