usaco Palindromic Squares

最新推荐文章于 2022-05-09 20:34:55 发布

原创最新推荐文章于 2022-05-09 20:34:55 发布 · 417 阅读

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#it #c语言

本文介绍了一个编程问题，即在给定的数制下找出其平方为回文数的所有整数，并提供了两种不同的C++实现方案。

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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)

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)

刚开始根本就没看清题，只考虑了10进制及以内的，后来发现是一直到20进制，之前在写时还在想怎么把数组和它的大小都传过来，真的好蠢！

改过后的代码（参考了一下网上的）：

/*
ID:
PROG: palsquare
LANG: C++
*/
#include<iostream>
#include<fstream>

using namespace std;
int N,l,L,x[50],y[50];
//用此函数转化进制
void trans(int n){
    int m;
    m=n;
    l=0;
    while(m){x[++l]=m%N;m/=N;}
    m=n*n;
    L=0;
    while(m){y[++L]=m%N;m/=N;}
}
//用此函数判断是否为回文
int judge(){
    for(int i=1;i<L/2+1;i++){
        if(y[i]!=y[L-i+1])return 0;
    }
    return 1;
}

int main(){
    freopen("palsquare.in","r",stdin);
    freopen("palsquare.out","w",stdout);
    cin>>N;
    //对于10进制及以内进制的数，eg.x[4]=4;c[4]=4;(不变)
    //对于10进制以上的数，eg.x[4]=10;c[10]=A,可正确表示
    char c[20]={'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J'};
    for(int i=1;i<=300;i++){
        trans(i);
        if(judge()){
            while(l){
                cout<<c[x[l--]];
            }
            cout<<" ";
            while(L){
                cout<<c[y[L--]];
            }
            cout<<endl;
        }
    }
    return 0;
}

之前的错代码（条理很不清）：

/*
ID:
PROG: palsquare
LANG: C++
*/
#include<iostream>
#include<fstream>

using namespace std;
int N;

int judge(int n){
    int t=n*n;
    int s[100],str[100];
    int rem,i=0;
    //把t转化成N进制，并存到数组s中
    do{
        rem=t%N;
        t/=N;
        s[i++]=rem;
    }while(t!=0);
    int flag=1;
    for(int j=0;j<i/2;j++){
        if(s[j]!=s[i-j-1]){flag=0;break;}
    }
    int b,k=0;
    if(flag){
        //如果n方是一个回文那么把n转化成N进制来输出；
        do{
        b=n%N;
        n/=N;
        str[k++]=b;
        }while(n!=0);
        while(k){
            k--;
            cout<<str[k];
        }
        cout<<" ";
        while(i){
            i--;
            cout<<s[i];
        }
        cout<<endl;
    }
}

int main(){
    freopen("palsquare.in 

","r",stdin);
    freopen("palsquare.out","w",stdout);
    cin>>N;
    int rem,j=0;
    for(int i=1;i<=300;i++){
        judge(i);
    }
    return 0;
}