[USACO] Palindromic Squares

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太COOL了，一次AC，虽然题简单，但还是好高兴。。。

此题主要就是考十进制转N进制，草纸上划划就出来了！

 

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/*ID:LANG: CTASK: palsquare*/#include <stdio.h>#include <stdlib.h>#include <string.h>#define SZ 20#define N 300char *num = "0123456789ABCDEFGHIJK";//2-20进制的所有字符void convert(int base, int n, char *p) {int i, k;i = 0;while(n > 0) {k = 0;while(n >= base) {n -= base;k++;}p[i++] = num[n];n = k;}p[i++] = '/0';}int isPalindromic(char *p) {int i, j, len;len = strlen(p);for(i = 0, j = len - 1; i < j; i++, j--)if(p[i] != p[j])return 0;return 1;}void reverse(char *p) {char c;int i, j, len;len = strlen(p);for(i = 0, j = len - 1; i < j; i++, j--) {c = p[i];p[i] = p[j];p[j] = c;}}int main() {FILE *fin = fopen("palsquare.in", "r");FILE *fout = fopen("palsquare.out", "w");char org[SZ], pal[SZ];int i, base;fscanf(fin, "%d", &base);for(i = 1; i <= N; i++) {convert(base, i, org);//把十进制的i转换成base进制的字符串convert(base, i * i, pal);if(isPalindromic(pal)) {//判断i的平方是不是回文数reverse(org);//反转i的字符串，因为存储的时候是逆序存储的fprintf(fout, "%s %s/n", org, pal);}}exit(0);} 

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分析的解法在求数的字符串表示时用了递归，这样存储时就是顺序存储的了，还有当i平方不是回文数的时候我也进行了i的转换，应该把i的转换写在里面，stupid:(

分析里有个写法s[len] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[n%b];就是求字符串对应下标的字符，然后进行赋值，以前没遇到这种写法，比较有意思。。。