This question requires the realization of a simple function for judging prime numbers, and use this function to verify Goldbach’s conjecture: any even number not less than 6 can be expressed as the sum of two odd prime numbers. A prime number is a positive integer that can only be divisible by 1 and itself. Note: 1 is not a prime number, 2 is a prime number.
Function interface definition:
int prime( int p );
void Goldbach( int n );
The function prime returns 1 when the user input parameter p is a prime number, otherwise it returns 0; the function Goldbach outputs the prime number decomposition of n according to the format "n=p+q", where p≤q is a prime number. And because such a decomposition is not unique (for example, 24 can be decomposed into 5+19, or it can be decomposed into 7+17), it is required to output the solution with the smallest p among all the solutions.
Sample referee test procedure:
#include <stdio.h>
#include <math.h>
int prime( int p );
void Goldbach( int n );
int main()
{
int m, n, i, cnt;
scanf("%d %d", &m, &n);
if ( prime(m) != 0 ) printf("%d is a prime number\n", m);
if ( m < 6 ) m = 6;
if ( m%2 ) m++;
cnt = 0;
for( i=m; i<=n; i+=2 ) {
Goldbach(i);
cnt++;
if ( cnt%5 ) printf(", ");
else printf("\n");
}
return 0;
}
/* 你的代码将被嵌在这里 */
Input sample:
89 100
Sample output:
89 is a prime number
90=7+83, 92=3+89, 94=5+89, 96=7+89, 98=19+79
100=3+97,
answer:
int prime( int p )
{
//参数p为素数时返回1，否则返回0
int i;
int flag = 1;
for (i = 2; i <= sqrt(p); i++) //详情见下面
{
if (p % i == 0)
{
flag = 0;
break;
}
}
if (flag == 1 && p != 1) return 1; //注意这里需要加一个条件 p 不等于 1 ，因为 1 不是素数
else return 0;
}
void Goldbach( int n )
{
//按照格式“n=p+q”输出n的素数分解，其中p≤q均为素数.要求必须输出所有解中p最小的解
int i; //循环变量
int j = 3; //分解的第一个素数 ，初值为 3 ，因为题目要求分解的数必须是素数
int flag = 1; //中介，用于判断是否是素数
int temp1 = 0, temp2 = 0; //用于存放 n 分解出来的两个素数
while ((temp1 + temp2) != n) //当temp1加temp2不等于n时循环
{
// flag = 1; //中介赋值为1，默认我们所要判断的数是素数
// for (i = 2; i <= sqrt(j); i++) //判断素数
// {
// if (j % i == 0)
// {
// flag = 0;
// break;
// }
// }
// if (flag == 1) temp1 = j; //如果是素数，把第一个素数给 temp1
if (prime(j) == 1) temp1 = j;
int t = n - temp1; //定义t变量，赋值为 n 减 temp1(因为其他的数，加起来不是没超过 n ，就是超过 n，减去就刚刚好) 例：87=90-3
// for (i = 2; i <= sqrt(t); i++) //判断第二个是不是素数
// {
// if (t % i == 0)
// {
// flag = 0;
// break;
// }
// }
// if (flag == 1) temp2 = t;
if (prime(t) == 1) temp2 = t;
j += 2; //每次 j 加二，因为初值为 3；题目要求分解出来的数是奇数的素数
}
printf("%d=%d+%d", n, temp1, temp2); //如果两个素数都找出来了就打印
}