If the number can be expressed as the sum of two prime numbers, the output shows the combination of the prime numbers. To perform this task, a user-defined function is created to check prime number. Integer as a Sum of Two Prime Numbers #include <stdio.h> int checkPrime(int n); int...
#include <stdio.h> // function to check number is prime or not // function will return 1 if number is prime int isPrime(int num) { int i; // loop counter // it will be 1 when number is not prime int flag = 0; // loop to check number is prime or not // we will check,...
,n); return 0; } int prime(int n) /* Function to check prime number */ { int i, flag=1; for(i=2; i<=n/2; ++i) if(n%i==0) flag=0; return flag; } 结果输出: Enter a positive integer: 34 34 = 3 + 31 34 = 5 + 29 34 = 11 + 23 34 = 17 + 17 7、用递归的...
// user-defined function to check prime number int checkPrimeNumber(int n) { int j, flag = 1; for (j = 2; j <= n / 2; ++j) { if (n % j == 0) { flag = 0; break; } } return flag; } Run Code OutputEnter two positive integers: 12 30 Prime numbers between 12 and...
prime = 0;/*表明 m 不是素数*/ if(prime)/*是素数*/ { printf("%6d", m); i ++; if(i % 5 == 0) printf("\n"); } } if(i % 5 != 0) printf("\n"); } 运行结果: 第二种采用筛选法来求素数: 使用筛选法求素数的基本思想是:把某一范围内的正整数按从小到大的顺序排列,宣布 ...
语法没有错误,内容上我按我的意思改了一下,你看这是不是你原来的目的:include <stdio.h> int main(){ int i, j ;int prime ;for( i=2 ; i<100 ; i++ ){ prime = 1 ;for( j=2 ; j
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(prime(x))wasobtainedfromthe/*functionresultsandoutputtheresultsofjudgingtrueorfalse*/Printf("No\t");ElsePrintf("yes\t");}Longintprime_super(longintx)todeterminewhetherthenumberoftheHan/**/superfunction{Inta,C,b;A=prime(x);/*calltodeterminethenumberHanprime,beadjustedreturn0or1*/functionB=...
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