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Anonymous users2024-02-07The above method needs to calculate the factor twice, the complexity increases, you can use the array to store the factor, the complete program is as follows.
#include
int main()
int i,j,sum=0;
int a[1000]=;The array holds the factor for(i=1; i<1000;i++)
sum=0;k=0;
for(j=1;j<=i/2;j++)
if (i%j==0) judgment factor.
a[k++]=j;Factors are stored in an array.
sum+=j;Find the sum of factors.
if(sum==i) if the sum of the factors is equal to itself printf("%5d",i);Output line breaks and the number itself printf(" its factors are :");
for(j=0;jreturn 0;
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Anonymous users2024-02-06int i,j,k,sum=0;
int a[50];
k=0;for(i=1;i<1000;i++)sum=0;
for(j=1;j<=i/2;j++)
if (i%j==0)
sum=sum+j;
if(sum==i)
a[k++]=i;
Then you can re-output the value of array a, and then make a loop output when you take its factor.
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Anonymous users2024-02-05Accumulate from 1 and validate with each additional value:
If the product is more than 1000, exit the loop, otherwise verify:
Whether the product is equal to the sum is the output.
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Anonymous users2024-02-04#include
#include
using namespace std;
int main()
int n;
int sum;
int i,j,k;
for(i = 1;i<=1000;i++)sum = 0;
n = i;
for(k = 2;kif(n%k == 0)
sum += k;
if(i == sum+1)
Cout: This method is a bit high, I don't know if this has a requirement, you try it.
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Anonymous users2024-02-03No, because there is not enough work.
while(j<=i) is an endless loop.
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