POJ 2413 How many Fibs?（高精度暴力）

How many Fibs?

Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 12308Accepted: 4404

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

题意

给你两个整数l,r，范围在10100内，问你在l~r内有多少个斐波那契数，定义Fib[1]=1,Fib[2]=2。

思路

虽然范围看起来十分大，但是由于斐波那契数增长十分快，所以到10100也就600个左右的数，所以直接暴力找就行了。（第一次手写高精1A..也算是一个成就吧QAQ）

Code

#pragma GCC optimize(3)#include<cstdio>#include<iostream>#include<cstring>#include<cctype>#include<string>#include<climits>#include<cmath>#include<algorithm>using namespace std;typedef long long ll;inline void readInt(int &x) {    x=0;int f=1;char ch=getchar();    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();    x*=f;}inline void readLong(ll &x) {    x=0;int f=1;char ch=getchar();    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();    x*=f;}/*================Header Template==============*/struct BigInt {    int a[105],len;    inline void read() {        memset(a,0,sizeof a);        char s[105];        scanf("%s",s+1);        len=strlen(s+1);        for(int i=1;i<=len;i++)            a[i]=s[len-i+1]-'0';    }    inline void init(ll x) {        memset(a,0,sizeof a);        int pos=0;        while(x) {            a[++pos]=x%10;            x/=10;        }        len=pos;    }    inline void getlimit() {        len=101;        for(int i=1;i<=100;i++)            a[i]=0;        a[101]=1;    }    inline void print() {        printf("Len = %d\n",len);        for(int i=len;i>=1;i--)            printf("%d",a[i]);        puts("");    }};inline BigInt operator + (const BigInt &a,const BigInt &b) {    BigInt c;    c.init(0LL);    c.len=max(a.len,b.len);    for(int i=1;i<=c.len;i++) {        c.a[i]+=a.a[i]+b.a[i];        if(c.a[i]>=10) {            c.a[i]-=10;            c.a[i+1]+=1;        }    }    if(c.a[c.len+1]!=0)        c.len++;    return c;}inline bool operator < (const BigInt &a,const BigInt &b) {//其实这里是小于等于    if(a.len<b.len)        return 1;    if(a.len>b.len)        return 0;    for(int i=a.len;i>=1;i--)        if(a.a[i]<b.a[i])            return 1;        else if(a.a[i]>b.a[i])            return 0;    return 1;}BigInt f[1010];BigInt l,r;int cnt=0;int main() {    f[1].init(1LL);    f[2].init(2LL);    for(cnt=3;;cnt++) {        f[cnt]=f[cnt-1]+f[cnt-2];        if(f[cnt].len>=101)            break;    }    while(1) {        l.read();        r.read();        if(l.len==1&&l.a[1]==0&&r.len==1&&r.a[1]==0)            break;        int ans=0;        for(int i=1;i<=cnt;i++) {            if(l<f[i]&&f[i]<r)                ans++;            if(r<f[i])                break;        }        printf("%d\n",ans);    }    return 0;}