uva 10183 - How Many Fibs?(斐波那契数）

题目连接：uva 10183 - How Many Fibs?

题目大意：给出a和b，求出a~b中有几个数时斐波那契数。

解题思路：模拟斐波那契数，找到临界的标号，相减的到答案。(大数，直接摘模板了，会比较长）

#include <stdio.h>
#include <string.h>
const int N = 105;

struct bign {
	int len, sex;
	int s[N];

	bign() {
		this -> len = 1;
		this -> sex = 0;
		memset(s, 0, sizeof(s));
	}

	bign operator = (const char *number) {
		int begin = 0;
		len = 0;
		sex = 1;
		if (number[begin] == '-') {
			sex = -1;
			begin++;
		}
		else if (number[begin] == '+')
			begin++;

		for (int j = begin; number[j]; j++)
			s[len++] = number[j] - '0';
	}

	bign operator = (int number) {
		char string[N];
		sprintf(string, "%d", number);
		*this = string;
		return *this;
	}

	bign (int number) {*this = number;}
	bign (const char* number) {*this = number;}

	bool operator < (const bign& b) const {
		if (len != b.len)
			return len < b.len;
		for (int i = 0; i < len; i++)
			if (s[i] != b.s[i])
				return s[i] < b.s[i];
		return false;
	}
	bool operator > (const bign& b) const { return b < *this; }
	bool operator <= (const bign& b) const { return !(b < *this); }
	bool operator >= (const bign& b) const { return !(*this < b); }
	bool operator != (const bign& b) const { return b < *this || *this < b;}
	bool operator == (const bign& b) const { return !(b != *this); }

	bign change(bign cur) {
		bign now;
		now = cur;
		for (int i = 0; i < cur.len; i++)
			now.s[i] = cur.s[cur.len - i - 1];
		return now;
	}

	bign operator + (const bign &cur){  
		bign sum, a, b;  
		sum.len = 0;
		a = a.change(*this);
		b = b.change(cur);

		for (int i = 0, g = 0; g || i < a.len || i < b.len; i++){  
			int x = g;  
			if (i < a.len) x += a.s[i];  
			if (i < b.len) x += b.s[i];  
			sum.s[sum.len++] = x % 10;  
			g = x / 10;  
		}  
		return sum.change(sum);  
	} 
};

int l, r;
char a[N], b[N];
bign f, s;
bign begin, end;

void solve() {
	int flag = 1;
	for (int i = 1; ; i++) {
		if (i % 2) {
			s = f + s;
			if (s >= begin && flag) {
				l = i;
				flag = 0;
			}
			if (s > end) {
				r = i;
				return ;
			}
		} else {
			f = f + s;	
			if (f >= begin && flag) {
				l = i;
				flag = 0;
			}
			if (f > end) {
				r = i;
				return ;
			}
		}
	}
}
int main () {
	while (scanf("%s%s", a, b) == 2) {
		if (a[0] == '0' && b[0] == '0') break;
		begin = a, end = b;
		f = 1, s = 0;
		solve();
		printf("%d\n", r - l);
	}
	return 0;
}