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USACO Section 1.2 Palindromic Squares

2019年04月07日 ⁄ 综合 ⁄ 共 1022字 ⁄ 字号 评论关闭

大意略。

/*
ID:g0feng1
LANG:C++
TASK:palsquare
*/

#include <iostream>
#include <fstream>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <map>
#include <algorithm>
#include <stack>
using namespace std;

ofstream fout("palsquare.out");
ifstream fin("palsquare.in");

int Base;

const int maxn = 110000;

int a[maxn], b[maxn];

void read_case()
{
	fin>>Base;
}

int check(int *a, int n)
{
	for(int i = 0; i < n/2; i++) if(a[i] != a[n-1-i]) return 0;
	return 1;
}

void print(int n, int base)
{
	stack<char> S;
	int s = n;
	while(s)
	{
		int t = s % base;
		s /= base;
		if(t <= 9)
		{
			char c = t+'0';
			S.push(c);
		}
		else if(t >= 10)
		{
			char c = t-10+'A';
			S.push(c);
		}
	}
	while(!S.empty()) { char c = S.top(); fout<<c; S.pop(); }
	fout<<" ";
	s = n*n;
	while(s)
	{
		int t = s % base;
		s /= base;
		if(t <= 9)
		{
			char c = t+'0';
			S.push(c);
		}
		else if(t >= 10)
		{
			char c = t-10+'A';
			S.push(c);
		}
	}
	while(!S.empty()) { char c = S.top(); fout<<c; S.pop(); }
	fout<<endl;
}

int change(int n, int base)
{
	int s = n;
	int len1 = 0, len2 = 0;
	s = n*n;
	while(s)
	{
		int t = s % base;
		s /= base;
		b[len2++] = t;
	}
	if(check(b, len2)) print(n, base);
}

void solve()
{
	read_case();
	for(int i = 1; i <= 300; i++)
	{
		change(i, Base);
	}
}

int main()
{
	solve();
	return 0;
}

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