This question requires the realization of a simple function to calculate the sum of integer factors, and use it to realize another function, outputting all the complete numbers between two positive integers m and n (0<m≤n≤10000). The so-called end number is that the number is exactly equal to the sum of factors other than itself. For example: 6=1+2+3, where 1, 2, 3 are factors of 6.
Function interface definition:
int factorsum( int number );
void PrintPN( int m, int n );
The function factorsum must return the sum of the factors of the int number; the function PrintPN must output the factor accumulation form of the factor accumulation of the given range [m, n] line by line, and each finished number occupies one line, and the format is "finished number" = Factor 1 + factor 2 +… + factor k", where the final number and factor are given in increasing order. If there is no perfect number in the given interval, output a line of "No perfect number".
Sample referee test procedure:
#include <stdio.h>
int factorsum( int number );
void PrintPN( int m, int n );
int main()
{
int m, n;
scanf("%d %d", &m, &n);
if ( factorsum(m) == m ) printf("%d is a perfect number\n", m);
if ( factorsum(n) == n ) printf("%d is a perfect number\n", n);
PrintPN(m, n);
return 0;
}
/* 你的代码将被嵌在这里 */
Input example 1:
6 30
Output sample 1:
6 is a perfect number
6 = 1 + 2 + 3
28 = 1 + 2 + 4 + 7 + 14
Input example 2:
7 25
Output sample 2:
No perfect number
answer:
int factorsum( int number )
{
//完数就是该数恰好等于除自身外的因子之和
int sum = 1; //定义sum为该数因子之和(由于1是每个数都有的因子，所以直接sum等一1)
int i; //循环变量 同时也是因子
for (i = 2; i < number; i++) //循环从2到number减一(每个数的因子都有1)
{
if (number % i == 0) sum += i; //如果能被2整除就是因子，累加y起来
}
return sum;
}
void PrintPN( int m, int n )
{
int flag = 0; //中介(如果有完数赋值为1，如果没有就是0)
int i, s; //循环变量 (同时i表示区间的每个数，s表示因子)
for (i = m;i <= n; i++) //区间循环
{
if (factorsum(i) == i) //如果调用上面的函数的求因子的和等于我们现在的数i，那么i就是完数
{
//如果 一个数的因子之和 等于这个数 那么该数就是完数
flag = 1; //有完数 flag赋值为1
printf("%d = 1", i); //由于每一个数都有因子1，所系先打印出来
for (s = 2; s < i; s++) //找因子
{
if (i % s == 0) printf(" + %d",s); //打印因子(格式需要找规律)
}
printf("\n"); //打印完一个完数需换行
}
}
if (flag == 0) printf("No perfect number"); //如果没有完数，打印相应信息
}
------------------------------Another way of writing---------------- ----------------------------------------
// int factorsum( int number )
// { //完数就是该数恰好等于除自身外的因子之和
// int sum = 1; //定义sum为该数因子之和(由于1是每个数都有的因子，所以直接sum等一1)
// int i; //循环变量 同时也是因子
// for (i = 2; i < number; i++) //循环从2到number减一(每个数的因子都有1)
// {
// if (number % i == 0) sum += i; //如果能被2整除就是因子，累加y起来
// }
// return sum;
// }
// void PrintPN( int m, int n )
// {
// int i, s; //循环变量(同时i表示区间的每个数，s表示因子)
// int cnt = 0; //中介，同时也表计数器，如果大于0就表示有完数
// for (i = m; i <= n; i++) //区间循环
// {
// int sum = 1; //一个数的因子之和
// for (s = 2; s < i; s++) //循环求一个数的因子之和
// {
// if (i % s == 0) sum+=s;
// }
// if (sum == i) //如果 一个数的因子之和 等于这个数 那么该数就是完数
// {
// cnt++; //计数器加一
// printf("%d = 1", i); //由于每一个数都有因子1，所系先打印出来
// for (s = 2; s < i; s++) //找因子
// { //6 = 1 + 2 + 3
// if (i % s == 0) printf(" + %d", s); //打印因子(格式需要找规律)
// }
// printf("\n"); //打印完一个完数需换行
// }
// }
// if (cnt == 0) printf("No perfect number"); //如果没有完数，打印相应信息
// }