筛选法求素数（Seeking prime number by screening method）.docVIP

筛选法求素数（Seeking prime number by screening method） Prime Palindromes 时间限制 (普通 / Java): 1000ms / 3000ms 运行内存限制: 65536kbyte 总提交: 60 测试通过: 7 描述 The Number 151 is a Prime Palindrome because it is both a prime number and a Palindrome (it is the same number when Read Forward as backward). Write a program that finds all Prime Palindromes in the range of two numbers a and B Group ( 5 = B = 100000000); both a and B are considered to be within the range. 输入 Only one line: Two integers, a and B 输出 The list of palindromic Primes in numerical order, one per line. 样例输入 5 500 样例输出 5 7 11 101 131 151 181 191 313 353 373 383 / * * * * * * * * * * * * * * * * * * * * * * * * * * * * Copyright (c) 2011, Hunan University of Science and Technology * Department of Computer Science and Engineering * All Rights Reserved * Completion date: 2011.12.11 * Author: Zhu Zhigang * Compiler: Visual Studio 2010 * File name: * * * * * * * * * * * * * * * * * * * * * * * * * * * * * / # iostream include # cMath include Using namespace STD; Judge (unsigned long int x) { Unsigned long I; For (i = 2; I = sqrt (double x); I + +) { If (x = = 0) return 0; } Return 1; } Int main () { Unsigned long A, B, PAL; Int D1, D2, D3, D4, I; CIN B; If ( 10) { For (i = 0; I ; I + +) { If (R b) return 0; If (Judge (i) cout I ENDL; } I = 11; Cout I ENDL; } For (d1 = 1 = 9; D1, D1 + +) For (D2 = 0 = 9; D2, D2 + +) { Pal = 100 * 10 * + D1 + D2 D1; If (PAL a) continue; If (PAL b) return 0; If (Judge (PAL) PAL) cout ENDL; } For (d1 = 1 = 9; D1, D1 + +) For (D2 = 0 = 9; D2, D2 + +) For (D3 = 0 = 9; D3; D3 + +) { Pal = 10000 * 1000 * D1 + D2 + D3 * 100 * 10 + D1 + D2; If (PAL a) continue; If (PAL b) return 0; If (Judge (PAL) PAL) cout ENDL; } For (d1 = 1 = 9; D1, D1 + +) For (D2 = 0 = 9; D2, D2 + +) For (D3 = 0 = 9; D3; D3 + +) For (D4 D4 = 0 = 9; D4 + +) { Pal = 1000000 + 100000 * * D1 D2 D3 + 10000 + D4 + * 1000 * 100 * 10 * D1 D2 D3 + +; If (PAL a) continue; If (PAL b) return 0; If (J