USACO 1.5 Prime Palindromes
1.生成回文数 （100000000以内）大概20000个
2.判断素数
生成回文数的算法思想：
分别生成数位是奇odd，和数位是偶even的回文数，
对于串1234, 翻转一下 4321，再接上就生成了一个回文数 ，odd:1234321 even:12344321
实现的方法可以多种多样。我用的是递归的方法。
代码
1 /*
2 ID: superbi1
3 LANG: C
4 TASK: pprime
5 */
6 #include <stdio.h>
7 #include <math.h>
8 #include <string.h>
9 #define NL 20000
10
11 int plin[NL], npl;
12
13 void genPrim(int high, int low, int val)
14 {
15 int I;
16 if (high < low) {
17 plin[npl++] = val;
18 return;
19 }
20 if (low == 1) I = 1;
21 else I = 0;
22 for (; I<=9; I++) {
23 if (high == low) genPrim(high-1, low+1, val+(int)pow(10, high-1)*I);
24 else genPrim(high-1, low+1, val+(int)pow(10, high-1)*I + (int)pow(10, low-1)*I);
25 }
26 }
27
28 int isPrim(int P)
29 {
30 int M = (int)sqrt(P);
31 int I;
32 for (I=2; I<=M; I++) {
33 if (P%I == 0) return 0;
34 }
35 return 1;
36 }
37
38 int main()
39 {
40 FILE *fin = fopen("pprime.in", "r");
41 FILE *fout = fopen("pprime.out", "w");
42 int I;
43 int a, b;
44 npl = 0;
45 for (I=1; I<=8; I++) {
46 genPrim(I, 1, 0);
47 }
48 fscanf(fin, "%d%d", &a, &b);
49 I = 0;
50 while (I<npl) {
51 if (plin[I] >= a) break;
52 I++;
53 }
54 while (I<npl && plin[I]<=b) {
55 if (isPrim(plin[I])) {
56 fprintf(fout, "%d\n", plin[I]);
57 }
58 I++;
59 }
60 return 0;
61 }
