7-1 Prime Day (20分)

The above picture is from Sina Weibo, showing May 23rd, 2019 as a very cool "Prime Day". That is, not only that the corresponding number of the date 20190523 is a prime, but all its sub-strings ended at the last digit 3 are prime numbers.

Now your job is to tell if a given date is a Prime Day.

Input Specification:

Each input file contains one test case. For each case, a date between January 1st, 0001 and December 31st, 9999 is given, in the format yyyymmdd.

Output Specification:

For each given date, output in the decreasing order of the length of the substrings, each occupies a line. In each line, print the string first, followed by a space, then Yes if it is a prime number, or No if not. If this date is a Prime Day, print in the last line All Prime!.

Sample Input 1:

20190523

Sample Output 1:

20190523 Yes0190523 Yes190523 Yes90523 Yes0523 Yes523 Yes23 Yes3 YesAll Prime!

Sample Input 2:

20191231

Sample Output 2:

20191231 Yes0191231 Yes191231 Yes91231 No1231 Yes231 No31 Yes1 No

题目限度：

题目粗心：

给定数字串，其子串定义为终点从0到最初地位，起点为最初地位的数字串，判断每一个子串是否为素数，如果是输入该数和Yes，否则输入该数和No，如果所有的子串都是素数，最初输入All Prime!

算法思路：

应用字符串s承受输出的数字，只有s长度大于0，就一直进行一下操作：

  • 1、将s转化为数字N，应用isPrime函数判断该数字是否为素数，如果是输入该数字和Yes，否则输入该数字和No，并应用allPrime记录不是所有子串都是素数。
  • 2、令s为下一个子串s = s.substr(1)，转1

    最初判断allPrime是否为true，如果是输入All Prime!

提交后果：

AC代码：

#include<cstdio>#include<vector>#include<cstring>#include<cmath>#include<string>#include<iostream>using namespace std;bool isPrime(int N){    if(N<=1) return false;    int sqrtn = (int)sqrt(1.0*N);    for (int i = 2; i <= sqrtn; ++i) {        if(N%i==0) return false;    }    return true;}int main(){    string s;    cin>>s;    bool allPrime = true;    while (s.size()>0){        int N = stoi(s);        if(isPrime(N)){            printf("%s Yes\n",s.c_str());        } else {            allPrime = false;            printf("%s No\n",s.c_str());        }        s = s.substr(1);    }    if(allPrime) printf("All Prime!");    return 0;}