疑問 - 請問該如何用C寫呢??

[rang0419]

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閱讀本主題的最佳解答
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An integer is said to be prime if it is divisible by only 1 and itself. For example, 2, 3, 5, and 7 are prime, but 4, 6, 8 and 9 are not.

1. Write a function that receives a number from the main function and determines if the number is prime.

2.Write the main function that
(a.)reads an integer n and uses the above function to determine and print all the prime numbers between 1 and n.

(b.)calulates the sum of all prime numbers between 1 and n.
A possibe run may look like:

Enter the number: 10
The prime numbers between 1 and 10 are:
2
3
5
7
The sum is 17.
Enter the number: 50
The prime numbers between 1 and 100 are:
2
3
5
7
11
13
17
19
23
29
31
37
41
43
47
The sum is 328.


可以參考一下嗎..........

[joebin]

#include<iostream>
using namespace std;
int main(){
  int s=0,n,m,i;  //s表總合,n表輸入的數字,m表要找的質數
   cout <<"please put in a num";
  cin >>n;
  if(n==1) cout <<"The sum is 0."<<endl;
  else{
    for(m=2;m<=n;m++)  //m由２到n
      for(i=2;i<=m;i++){
        if(m%i==0&&i!=m) break;  //當m不為質數時跳出
             if(i==m){
          cout <<m<<endl;  //m為質數,s+m=s
          s+=m;
        }
      }
  }
  cout <<"The sum is "<<s<<"."<<endl;
  system("pause");
  return 0;
}

還有The prime numbers between 1 and 100 are

這句有誤,應該是The prime numbers between 1 and 50 are

[rang0419]

非常感謝你的解答喔!!

很詳細又有註解~~~

我現在就來試試看..........

感謝捏 有問題我再問你!!

[rang0419]

還有錯誤喔 不會吧 囧........

[joebin]

你看一下他題目要求的輸出方式

這邏輯應該無誤才對.......