POJ 2506 Tiling 高精度

来源：互联网发布：营销软件app 编辑：程序博客网时间：2024/04/29 22:59

题目大意：给出一个2*n的条形区域，问用2*1和2*2的方格一共有多少种摆放的方法。

思路：f[i] = f[i - 1] + f[i - 2] * 2

写一个高精度加法就可以了。

CODE：

#include <cstdio>#include <cstring>#include <iomanip>#include <iostream>#include <algorithm>#define MAX 260#define BASE 1000using namespace std;struct BigInt{int num[MAX],len;BigInt(int _ = 0) {memset(num,0,sizeof(num));if(_) {num[1] = _;len = 1;}elselen = 0;}BigInt operator +(const BigInt &a)const {BigInt re;re.len = max(len,a.len);int temp = 0;for(int i = 1; i <= re.len; ++i) {re.num[i] = num[i] + a.num[i] + temp;temp = re.num[i] / BASE;re.num[i] %= BASE;}if(temp)re.num[++re.len] = temp;return re;}}f[MAX];ostream &operator <<(ostream &os,const BigInt &a) {os << a.num[a.len];for(int i = a.len - 1; i; --i)os << fixed << setfill('0') << setw(3) << a.num[i];return os;}int main(){f[0] = BigInt(1);f[1] = BigInt(1);f[2] = BigInt(3);for(int i = 3; i <= 250; ++i)f[i] = f[i - 1] + f[i - 2] + f[i - 2];int x;while(scanf("%d",&x) != EOF)cout << f[x] << endl;return 0;}

0 0