USACO 1.2.5 Dual Palindromes (dualpal)

1、從S開始枚舉每個十進位數，判斷其對應的B進位是否迴文數。

/*<br />ID: gengjia1<br />LANG: C<br />TASK: dualpal<br />*/<br />#include <stdio.h><br />#include <stdlib.h><br />//#define NDEBUG<br />#include <assert.h><br />int N, S;<br />const char *num = "0123456789";<br />int main(void) {<br />FILE *fin = fopen ("dualpal.in", "r");<br />FILE *fout = fopen ("dualpal.out", "w");<br />int pal;<br />int tmp;<br />int B;<br />char b[33];<br />int count;<br />int i, j, k;<br />fscanf (fin, "%d", &N);<br />assert(N >= 1);<br />assert(N <= 15);<br />fscanf (fin, "%d", &S);<br />assert(S > 0);<br />assert(S < 10000);<br />i = 0;<br />while(i < N)<br />{<br />S += 1;<br />count = 0;<br />for(B = 2; B < 11; B++)<br />{<br />tmp = S;<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';<br />//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 />}<br />if(pal == 1)<br />{<br />count += 1;<br />if(count >= 2)<br />{<br />fprintf(fout, "%d/n", S);<br />i += 1;<br />break;<br />}<br />}<br />}<br />}<br />fclose(fin);<br />fclose(fout);<br />exit(0);<br />}<br />