USACO SEC.1.2 No.4 Palindromic Squares

题意：输入一个进制b（[2,20]），确定从[1,300]中哪些数的平方在该进制下是回文数

解法：核心部分是将十进制数转换为任意进制数，除x取余，逆序排列

/*
ID: lsswxr1
PROG: palsquare
LANG: C++
*/
#include <iostream>
#include <vector>
#include <map>
#include <list>
#include <set>
#include <deque>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <cstdio>
#include <iomanip>
#include <cmath>
#include <cstdio>
#include <string>
#include <fstream>
using namespace std;

///宏定义
const int  INF = 1000000000;
const int MAXN = 15;
const int maxn = MAXN;
///全局变量 和 函数

#define USACO
#ifdef USACO
#define cin fin
#define cout fout
#endif
//////////////////////////////////////////////////////////////////////////
int b;
stack<int> stk, origin, square;
queue<int> que;
char str[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
int main()
{


#ifdef USACO    
    ofstream fout ("palsquare.out");
    ifstream fin ("palsquare.in");
#endif    
    ///变量定义
    while(cin >> b)
    {
        for (int i = 1; i <= 300; i++)
        {
            bool flag = true;
            while (!origin.empty())
                origin.pop();
            while (!square.empty())
                square.pop();
            while (!stk.empty())
                stk.pop();
            while (!que.empty())
                que.pop();
            int k = i * i;
            int o = i;
            while (o != 0)
            {
                int yuu = o % b;
                origin.push(yuu);
                o /= b;
            }
            while (k != 0)
            {
                int yu = k % b;
                stk.push(yu);
                que.push(yu);
                square.push(yu);
                k /= b;
            }
            while ((!stk.empty()) && (!que.empty()))
            {
                int f1, f2;
                f1 = stk.top();
                f2 = que.front();
                if (f1 != f2)
                {
                    flag = false;
                    break;
                }
                stk.pop();
                que.pop();
            }
            if (flag)
            {
                while (!origin.empty())
                {
                    int index = origin.top();
                    cout << str[index];
                    origin.pop();
                }
                cout << " ";
                while (!square.empty())
                {
                    int index = square.top();
                    cout << str[index];
                    square.pop();
                }
                cout << endl;
            }
        }
    }

    ///结束
    return 0;
}

 

官方的解法：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>
#include <math.h>

/* is string s a palindrome? */
int
ispal(char *s)
{
    char *t;

    t = s+strlen(s)-1;
    for(t=s+strlen(s)-1; s<t; s++, t--)
        if(*s != *t)
        return 0;

    return 1;
}

/* put the base b representation of n into s: 0 is represented by "" */
// 进制转换的递归写法，可以借鉴和学习
void
numbconv(char *s, int n, int b)
{
    int len;

    if(n == 0) {
    strcpy(s, "");
    return;
    }

    /* figure out first n-1 digits */
    numbconv(s, n/b, b);

    /* add last digit */
    len = strlen(s);
    s[len] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[n%b];
    s[len+1] = '\0';
}

void
main(void)
{
    char s[20];
    char t[20];
    int i, base;
    FILE *fin, *fout;

    fin = fopen("palsquare.in", "r");
    fout = fopen("palsquare.out", "w");
    assert(fin != NULL && fout != NULL);

    fscanf(fin, "%d", &base);
    for(i=1; i <= 300; i++) {
    numbconv(s, i*i, base);
    if(ispal(s)) {
        numbconv(t, i, base);
        fprintf(fout, "%s %s\n", t, s);
    }
    }
    exit(0);
}