USACO:Palindromic Squares;回文判断+进制转化

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/*TASK:palsquareLANG:C++ID:huibaochen*//*一题：给你一个数n代表进制，算出从1到300的数的平方转化为n进制后是否进回文序列，是的话输出这个是的n进制和这个数的平方的n进制*/#include <iostream>#include <stdio.h>#include <string.h>int m, n, len, num, r;char str[1000 +5], ch[100 + 5];using namespace std;void yes(int m){    num = m * m;    len = 0;    while(num){        r = num % n;        num = num / n;        if(r >= 0 && r <= 9)            str[len++] = r + 48;        else            str[len++] = r + 'A' - 10;    }    for(int i = 0; i < len / 2; i++){        if(str[i] != str[len - 1 - i])            return;    }    len = 0;    while(m){       r = m % n;       m = m / n;       if(r >= 0 && r <= 9)            ch[len++] = r + 48;       else            ch[len++] = r + 'A' - 10;    }    for(int i = 0; i < len / 2; i++){        swap(ch[i], ch[len - i - 1]);    }    printf("%s %s\n", ch, str);}int main(){    freopen ("palsquare.in", "r", stdin);    freopen ("palsquare.out", "w", stdout);    scanf("%d", &n);    for(int i = 1; i <= 300; i++){        yes(i);    }    return 0;}

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