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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;
}
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