跪求，改下两段小程序

[huiguipingdan]

求716539的质因子？

#include<stdio.h>
#include<math.h>
void main()
{
	int i,j,n=716539;
	for(i=2;i<=sqrt(n);i++)
	
	if(n%i==0)
	{
	for(j=2;j<=sqrt(i);j++)
	if(i%j==0)
	printf("zhi yin zi: %d\n",i);
	}	
}

当A*B=716699时，A+B之和最小为？

#include<stdio.h>
void main()
	{
long a,b,min=716699;
for(a=0;a<716699;a++)
 for(b=0;b<716699;b++)
	if(b*a==716699)
	min=a+b>min?min:a+b;
	printf("a=%ld ,b=%ld,\n",a,b);	
	}

[pamtry]

Num 1:

#include<stdio.h>
#include<math.h>
void main()
{
    int i,j,n=716539;
    for(i=2;(double)i <= sqrt((double)n);i++)
    {
        if(0 == n % i)
        {
            for(j=2;(double)j <= sqrt((double)i);j++)
                if(i%j==0) break;
			if((double)j > sqrt((double)i))
				printf("zhi yin zi: %d\n",i);
        }
    }    
}

[delphiwcdj]

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main()
{
	int n, i;
	n = 716539;

	printf ("716539 = ");
	for (i=2;i<=n;i++)
	{
		while (n%i==0)
		{
			printf ("%d*",i);
			n=n/i;

		}
	}   
	printf ("\b");

	system("pause");
	return 0;
}

[pamtry]

Num 2:

#include<stdio.h>
void main()
{
	long a,b,min=716699;
	for(a=1;a<716699;a++)
	{
		if(0 != 716699 % a)
			break;
		b = 716699 / a;
		if((a + b) < min)
			min = a + b;
	}
	printf("Min A+B is: %d", min);
}

[awsqsh]

2楼正确 。、。。

[pamtry]

[Quote=引用 2 楼 delphiwcdj 的回复:]

C/C++ code

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main()
{
int n, i;
n = 716539;

printf ("716539 = ");
for (i=2;i<=n;i++)
{
while (n%i==0)
……
[/Quote]

+1,忘记质因数分解后对原数范围进行缩小，这种方式对于可以质因数分解的数，效率会高很多

[l369294289]

个人觉得第二题可以这样，求出716699的所有因子用数组存起来，然后找出最中间那个因子A，用716699除以A得B，这样的A+B应该最小。。。

[delphiwcdj]

2,

#include<stdio.h>

int main()
{
	long i, j, min = 716699, a, b;

	for (i = 1; i < 716699; ++i)
	{
		if (716699%i == 0)
		{
			j = 716699 / i;
			if (i+j < min)
			{
				a = i;
				b = j;
				min = i+j;
			} 
		}
	}
	printf("%ld = %ld+%ld\n", min, a, b);    

	return 0;
}

[delphiwcdj]

#include<stdio.h>
#define N 716699

int main()
{
	long i, j, min = 716699, a, b;

	for (i = 1; i < N; ++i)
	{
		if (N%i == 0)
		{
			j = N / i;
			if (i+j < min)
			{
				a = i;
				b = j;
				min = i+j;
			} 
		}
	}
	printf("%ld = %ld*%ld\n", N, a, b);    
	printf("%ld = %ld+%ld\n", min, a, b);   

	return 0;
}
/*
output:

716699 = 563*1273
1836 = 563+1273

*/

[沭水河畔]

貌似我的代码效率要高很多：

#include <stdio.h>
#include <math.h>
#define N 716699
#define true 1

int main(void)
{
	int a=(int)sqrt(N);
	while(true)
	{
		if(N%a==0)
		{
			printf("%d+%d=%d\n",a,N/a,(a+N/a));
			break;
		}
		a--;
	}
	return 0;
}	

[noteboom]

716539=83*89*97

A+B=7484

[huiguipingdan]

[Quote=引用 1 楼 pamtry 的回复:]

Num 1:
C/C++ code
#include<stdio.h>
#include<math.h>
void main()
{
int i,j,n=716539;
for(i=2;(double)i <= sqrt((double)n);i++)
{
if(0 == n % i)
{
for(j=2;(do……
[/Quote]
貌似我的程序只要加句if(i%j==0) break;