uva 10334 - Ray Through Glasses(斐波那契数）

题目链接：uva 10334 - Ray Through Glasses

题目大意：在2块玻璃中反射k次的光线条数。

解题思路：斐波那契数列。(大数）

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;const int N = 1005;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);  } void delZore() {// 删除前导0.bign now = change(*this);while (now.s[now.len - 1] == 0 && now.len > 1) {now.len--;}*this = change(now);}void put() {    // 输出数值。delZore();if (sex < 0 && (len != 1 || s[0] != 0))cout << "-";for (int i = 0; i < len; i++)cout << s[i];}};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 ;}}}}bign num[N];void init() {num[0] = 1;num[1] = 2;for (int i = 2; i < N; i++)num[i] = num[i - 1] + num[i - 2];}int main () {init();int n;while (scanf("%d", &n) == 1) {num[n].put();printf("\n");}return 0;}