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要求实现一个计算整数因子和的简单函数，并利用其实现另一个函数，输出两正整数m和n（0<m≤n≤10000）之间的所有完数。所谓完数就是该数恰好等于除自身外的因子之和。例如：6=1+2+3，其中1、2、3为6的因子。
怎么找公因数简单一些，循环太多次导致代码处理时间长
怎么找公因数简单一些，循环太多次导致代码处理时间长
#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;
}
int factorsum( int number ) //判断是否为完数
{
int i,j;
int m = 1;
int n,num = 0;
int a[100] = {0};
a[0] = 1;
if(number == 0)
return -1;
if(number == 1)
return 1;
for(i = 2; i < number; i ++ )
{
for(j = i + 1; j < number; j ++)
{
if(i * j == number)
{
a[m] = i;
m ++;
a[m] = j;
m ++;
}
}
}
for(n = 0; n < 100; n ++)
{
if(a[n] == 0)
break;
num = num + a[n];
}
if(num == number)
return num;
else
return 0;
}
void PrintPN( int m, int n ) //找出m到n之间的完数
{
int q;
int i,j;
int num = 0;
for(int k = m; k <= n; k++)
{
if(factorsum(k) == k)
{
num = 1;
int p = 1;
int a[100] = {0};
a[0] = 1;
for(i = 2; i < k; i ++ )
{
for(j = i + 1; j < k; j ++)
{
if(i * j == k)
{
a[p] = i;
p ++;
a[p] = j;
p ++;
}
}
}
//排序
int b;
for(b = 0; a[b] != 0; b++)
{
for(int c = b + 1; a[c] != 0; c ++)
{
if(a[b] > a[c])
{
int key;
key = a[b];
a[b] = a[c];
a[c] = key;
}
}
}
printf("%d = ",k);
for( q = 0; q < 100; q ++)
{
if(a[q] == 0)
{
printf("\n");
break;
}
if(q == 0)
printf("%d",a[q]);
else
printf(" + %d",a[q]);
}
}
}
if(num == 0)
printf("No perfect number\n");
}
...全文
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forever74 2021-02-23
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要那样输出的话可以参考这个：
#include <stdio.h>
#include <math.h>
int factorsum(int n, int* fac) //除了计算因子和，还在数组fac里给出因子列表，以-1表示结束
{
int i, sum = 1, k, *p = fac + 1, half[40], j = 0;
if (n == 1)return 0;
for (i = 2; i <= (int)(sqrt(n) + 0.001); i++) //从2到根号n测试
{
if (n%i == 0) //每次整除产生两个因子
{
sum += (*p++ = i); //小的那个直接升序记入数组
if (i < (k=n/i)) //除非 i==k 这次只产生1个因子
sum += (half[j++] = k); //大的那个降序另存
}
}
while (--j >= 0) //把较大的一半因子逆序抄过去
{
*p++ = half[j];
}
*p = -1; //表示结束
return sum;
}
void PrintPN(int m, int n)
{
int i, fac[70] = { 1 };
for (i = m; i <= n; i++)
{
if (i == factorsum(i, fac))
{
int *p = fac + 1;
printf("%d=1", i);
while (*p > 0)
{
printf("+%d", *p++);
}
printf("\n");
}
}
}
int main()
{
int m, n;
scanf("%d%d", &m, &n);
PrintPN(m, n);
return 0;
}
qzjhjxj 2021-02-23
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判断是否为完数的函数再优化一点点：
int factorsum( int n)//判断是否为完数
{
int k=2,s=1;
while(k <= n/2){
if(n%k==0)s+=k;
k++;
}
if(s==n && n!=1) return 1;
return 0;
}
自信男孩 2021-02-23
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#include <stdio.h>
int factorsum( int number );
void PrintPN( int m, int n );
static void swap(int *a, int *b);
static void print_perfect(int m, int n);
int main()
{
int m, n;
scanf("%d %d", &m, &n);
if (m > n) {
swap(&m, &n);
}
print_perfect(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;
}
static void swap(int *a, int *b)
{
int tmp;
tmp = *a;
*a = *b;
*b = tmp;
}
static void print_perfect(int m, int n)
{
int i;
for (i = m; i < n; i++) {
if (i == 1)
continue;
if (factorsum(i)) {
printf("%d is a perfect number!\n", i);
}
}
}
int factorsum( int number ) //判断是否为完数
{
int i;
int sum = 1;
for (i = 2; i < number / 2 + 1; i++) {
if (number % i == 0)
sum += i;
}
if (sum == number)
return 1;
return 0;
/*
int i,j;
int m = 1;
int n,num = 0;
int a[100] = {0};
a[0] = 1;
if(number == 0)
return -1;
if(number == 1)
return 1;
for(i = 2; i < number; i ++ )
{
for(j = i + 1; j < number; j ++)
{
if(i * j == number)
{
a[m] = i;
m ++;
a[m] = j;
m ++;
}
}
}
for(n = 0; n < 100; n ++)
{
if(a[n] == 0)
break;
num = num + a[n];
}
if(num == number)
return num;
else
return 0;
*/
}
void PrintPN( int m, int n ) //找出m到n之间的完数
{
int q;
int i,j;
int num = 0;
for(int k = m; k <= n; k++)
{
if(factorsum(k) == k)
{
num = 1;
int p = 1;
int a[100] = {0};
a[0] = 1;
for(i = 2; i < k; i ++ )
{
for(j = i + 1; j < k; j ++)
{
if(i * j == k)
{
a[p] = i;
p ++;
a[p] = j;
p ++;
}
}
}
//排序
int b;
for(b = 0; a[b] != 0; b++)
{
for(int c = b + 1; a[c] != 0; c ++)
{
if(a[b] > a[c])
{
int key;
key = a[b];
a[b] = a[c];
a[c] = key;
}
}
}
printf("%d = ",k);
for( q = 0; q < 100; q ++)
{
if(a[q] == 0)
{
printf("\n");
break;
}
if(q == 0)
printf("%d",a[q]);
else
printf(" + %d",a[q]);
}
}
}
if(num == 0)
printf("No perfect number\n");
}供参考～
qzjhjxj 2021-02-23
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修正楼上第10行：
for(int i=m;i<=n;i++){qzjhjxj 2021-02-23
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楼主考虑的太复杂了，供参考：
#include <stdio.h>
int factorsum( int n );//判断是否为完数
int main()
{
int m, n;
scanf("%d %d", &m, &n);
if(m<=0 || n>10000) return -1;
for(int i=m;i<n;i++){
if(factorsum(i))printf("%d is a perfect number\n", i);
}
return 0;
}
int factorsum( int n)
{
int k=1,s=0;
while(k <= n/2){
if(n%k==0) s+=k;
k++;
}
if(s==n) return 1;
return 0;
}
赵4老师 2021-02-23
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随便搜个对正整数分解因数的代码
