Palindromic Squares（模拟）（USACO）

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本文介绍了一个算法问题，即在给定的数制下找出所有其平方为回文数的整数（1 ≤ N ≤ 300）。文章提供了一段C++代码，实现了将十进制数转换为指定进制并判断该数是否为回文数的功能。

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http://train.usaco.org/usacoprob2?S=palsquare&a=wnFTyrFubcM
Palindromic Squares
Rob Kolstad
Palindromes are numbers that read the same forwards as backwards. The number 12321 is a typical palindrome.

Given a number base B (2 <= B <= 20 base 10), print all the integers N (1 <= N <= 300 base 10) such that the square of N is palindromic when expressed in base B; also print the value of that palindromic square. Use the letters ‘A’, ‘B’, and so on to represent the digits 10, 11, and so on.

Print both the number and its square in base B.

PROGRAM NAME: palsquare

INPUT FORMAT

A single line with B, the base (specified in base 10).
SAMPLE INPUT (file palsquare.in)

10
OUTPUT FORMAT

Lines with two integers represented in base B. The first integer is the number whose square is palindromic; the second integer is the square itself. NOTE WELL THAT BOTH INTEGERS ARE IN BASE B!
SAMPLE OUTPUT (file palsquare.out)

1 1
2 4
3 9
11 121
22 484
26 676
101 10201
111 12321
121 14641
202 40804
212 44944
264 69696

思考：
1、看题一定要看全

/*
ID: emaguo1
PROG: palsquare
LANG: C++11
*/
#include <iostream>
#include <fstream>
#include <string>
#include <cstring>
#include <map>
#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn = 20;

void ten2base(char *a, int b, int base)
{
    int i = 0;
    while (b) {
        int c = b % base;
        if (c < 10) {
            a[i++] = c + '0';
        } else {
            a[i++] = c - 10 + 'A';
        }
        b /= base;
    }
}

bool ispalindrome(char *a, int b, int base)
{
    ten2base (a, b, base);
    int l = strlen (a);
    for (int i = 0; i < l / 2; i++) {
        if (a[i] != a[l - 1 - i]) return false;
    }
    return true;
}

int main() {
    freopen("palsquare.in","r",stdin);
    freopen("palsquare.out","w",stdout);
    int base;
    scanf ("%d", &base);
    char a[maxn], b[maxn];
    memset (a, '\0', sizeof(a));
    memset (b, '\0', sizeof(b));
    for (int i = 1; i <= 300; i++) {
        int square = i * i;
        if (ispalindrome (a, square, base)) {
            ten2base (b, i, base);
            //cout << b << endl;
            int l = strlen (b);
            for (int i = l - 1; i >= 0; i--) {
                cout << b[i];
            }
            cout << " " << a << endl;
        }
    }
    return 0;
}