USACO Section 1.5 Prime Palindromes
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题目描述
151号是一个主要的回文,因为它是素数和回文(当向前读取时,它是相同的数字)。 编写一个程序,找到两个提供的数字a和b(5 <= a <b <= 100,000,000)范围内的所有主回文; a和b都被认为在该范围内。程序名称:pprime
输入格式
行1:两个整数,a和b
输入 (file pprime.in)
5 500
输出格式
数字序列中的回文素数列表,每行一个
输出(file pprime.out)
5711101131151181191313353373383提示(仔细使用它们)!
提示1 提示2
解题代码
/*ID: 15189822PROG: pprimeLANG: C++*/#include<iostream>#include<cmath>#include<fstream>using namespace std;ifstream fin("pprime.in");ofstream fout("pprime.out");const int N = 1000;long x[N+1];int cnt=0;long a,b;bool is_prime(long n){ if (n<a||n>b){ return false; } if (n==0||n==1) return false; if (n==2||n==3) return true; long i; for (i=2;i<=sqrt(n);i++){ if (n%i==0) return false; } return true;}void hui_8(){ int x1,x2,x3,x4; for (x1=1;x1<=9;x1+=2){ for (x2=0;x2<=9;x2++){ for (x3=0;x3<=9;x3++){ for (x4=0;x4<=9;x4++){ long n=(x1*1000+x2*100+x3*10+x4)*10000+x4*1000+x3*100+x2*10+x1; if (is_prime(n)) fout<<n<<endl; } } } }}void hui_7(){ int x1,x2,x3,x4; for (x1=1;x1<=9;x1+=2){ for (x2=0;x2<=9;x2++){ for (x3=0;x3<=9;x3++){ for (x4=0;x4<=9;x4++){ long n=(x1*1000+x2*100+x3*10+x4)*1000+x3*100+x2*10+x1; if (is_prime(n)) fout<<n<<endl; } } } }}void hui_6(){ int x1,x2,x3; for (x1=1;x1<=9;x1+=2){ for (x2=0;x2<=9;x2++){ for (x3=0;x3<=9;x3++){ long n=(x1*100+x2*10+x3)*1000+x3*100+x2*10+x1; if (is_prime(n)) fout<<n<<endl; } } }}void hui_5(){ int x1,x2,x3; for (x1=1;x1<=9;x1+=2){ for (x2=0;x2<=9;x2++){ for (x3=0;x3<=9;x3++){ long n=(x1*100+x2*10+x3)*100+x2*10+x1; if (is_prime(n)) fout<<n<<endl; } } }}void hui_4(){ int x1,x2; for (x1=1;x1<=9;x1+=2){ for (x2=0;x2<=9;x2++){ long n=(x1*10+x2)*100+x2*10+x1; if (is_prime(n)) fout<<n<<endl; } }}void hui_3(){ int x1,x2; for (x1=1;x1<=9;x1+=2){ for (x2=0;x2<=9;x2++){ long n=(x1*10+x2)*10+x1; if (is_prime(n)) fout<<n<<endl; } }}void hui_2(){ int x1; for (x1=1;x1<=9;x1+=2){ long n=x1*10+x1; if (is_prime(n)) fout<<n<<endl; }}void hui_1(){ for (int i=2;i<=9;i++){ if (is_prime(i)) fout<<i<<endl; } }int main(){ cnt=0; fin>>a>>b; hui_1(); hui_2(); hui_3(); hui_4(); hui_5(); hui_6(); hui_7(); hui_8(); return 0;}
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