USACO 1.2.4 Palindromic Squares(palsquare)

1、唯一知識點：進位轉換

int 位元=0;<br />while(j不為零)<br />{<br /> b[位元]=j mod 進位 ;<br /> j = j / 進位;<br /> 位元=位元+1;<br />}  

2、思路：窮舉[1,300]所有平方數，轉進位，判斷是否迴文數。注意原數逆序輸出，迴文數不需要。

/*<br />ID: gengjia1<br />LANG: C<br />TASK: palsquare<br />*/<br />#include <stdio.h><br />#include <stdlib.h><br />//#define NDEBUG<br />#include <assert.h></p><p>#define N 300<br />int B;<br />char num[20] = {'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J'};<br />char a[10] = {0};<br />char b[15] = {0};</p><p>int main(void) {<br />FILE *fin = fopen ("palsquare.in", "r");<br />FILE *fout = fopen ("palsquare.out", "w");</p><p>int i, j, k, p;<br />int pal;<br />int tmp;<br />char t;</p><p>fscanf (fin, "%d", &B);<br />assert(B >= 2);<br />assert(B <= 20);</p><p>for(i = 1; i <= N; i++)<br />{<br />tmp = i * i;<br />pal = 1;<br />j = 0;<br />while(tmp != 0)<br />{<br />b[j] = num[tmp % B];<br />tmp = tmp / B;<br />j += 1;<br />}<br />b[j] = '/0';</p><p>//判斷迴文<br />for(k = 0; k < j / 2; k++)<br />{<br />if(b[k] != b[j-k-1])<br />{<br />pal = 0;<br />break;<br />}<br />}</p><p>if(pal == 1)<br />{<br />tmp = i;<br />k = 0;<br />while(tmp != 0)<br />{<br />a[k] = num[tmp % B];<br />tmp = tmp / B;<br />k += 1;<br />}<br />for(p = 0; p < k/2; p++)<br />{<br />t = a[p];<br />a[p] = a[k-p-1];<br />a[k-p-1] = t;<br />}<br />a[k] = '/0';<br />fprintf(fout, "%s %s/n", a, b);<br />}<br />}</p><p>fclose(fin);<br />fclose(fout);<br />exit(0);<br />}<br />