c++
//#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
inline int read()
{
char ch=getchar();
int a=0,t=1;
while(ch<'0'||ch>'9')
{
if(ch=='-') t=-1;
ch=getchar();
}
while(ch<='9'&&ch>='0')
{
a=a*10+ch-'0';
ch=getchar();
}
return a*t;
}
int main()
{
int n;
//freopen("","r",stdin);freopen("","w",stdout);
n=read();
for(int i=2;i<=sqrt(n);++i)
{
if(n%i==0)
{
printf("%d",n/i);
return 0;
}
}
//fclose(stdin);fclose(stdout);
return 0;
}
//
如果一个数是两个质数乘积，那么一定不是两个非质数乘积

#include<iostream>
using namespace std;
int pd(int n)
{
int i;
if(n<=1)
return 0;
for(i=2;i*i<=n;i++)
if(n%i==0)
return n/i;
}
int main()
{
int n;
cin>>n;
cout<<pd(n)<<endl;
return 0;
}

记录信息
评测状态 Accepted
题目 P1786 质因数分解
递交时间 2013-11-20 22:52:46
代码语言 C
评测机 上海红茶馆
消耗时间 0 ms
消耗内存 540 KiB
评测时间 2013-11-20 22:52:47

评测结果
编译成功

foo.c: In function 'main':
foo.c:9:1: warning: control reaches end of non-void function [-Wreturn-type]

测试数据 #0: Accepted, time = 0 ms, mem = 536 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 536 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 540 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 536 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 540 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 536 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 540 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 536 KiB, score = 10

测试数据 #8: Accepted, time = 0 ms, mem = 536 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 540 KiB, score = 10

Accepted, time = 0 ms, mem = 540 KiB, score = 100

代码
#include <stdio.h>
int main (){
long n,i;
scanf ("%ld",&n);
for (i=2;i<n;i++)
{if (n%i==0) {printf ("%ld\n",n/i);return 0;}
}
//若vijos给AC了,vijos是猪头.
}
用户信息
lemontree712

题目操作
查看 P1786
查看本题题解
查看本题递交记录
我的记录
11-20 Accepted
11-20 Accepted
​结论:……

太简单了，无话可说，说白了就是分解因数，题意说了两个不同素数，直接分解即可
#include<cstdio>
#include<iostream>

using namespace std;

int main(){
int n,i;
scanf("%d",&n);//读取;
for(i=2;i<n;i++)
{
if(n%i==0)//因数分解
{
printf("%d\n",n/i);//输出较大的
break;//得到退出
}
}
return 0;
}

var n,i:longint;
begin
read(n);
for i:=2 to trunc(sqrt(n)) do
  if n mod i=0 then begin writeln(n div i);halt;end;
end.

只要找小的，再出一下就好啦~~~直接暴力!!!开始从大搜，60分+超时，从小的搜再i++就可以了~~~
测试数据 #0: Accepted, time = 0 ms, mem = 560 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 556 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 556 KiB, score = 10

测试数据 #3: Accepted, time = 15 ms, mem = 560 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 560 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 560 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 560 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 556 KiB, score = 10

测试数据 #8: Accepted, time = 15 ms, mem = 556 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 556 KiB, score = 10
再来代码~~~
#include<iostream>
#include<cmath>
using namespace std;
bool pd(long int n)
{
if(n==0||n==1)
return false;
if(n==2)
return true;
for(long int i=2;i<=sqrt(n);i++)
{
if(n%i==0)
return false;
}
return true;
}
int main()
{
long int n,mi;
cin>>n;
for(long int i=2;i<=sqrt(n);i++)
{
bool check=pd(i);
if(check==true&&n%i==0)
{
mi=i;
break;
}
}
cout<<n/mi;
return 0;
}

#include <stdio.h>
#include <math.h>
int main(){
    freopen("prime.in","r",stdin);
    freopen("prime.out","w",stdout);
    int n;
    scanf("%d", &n);
    for (int i=2;i<=sqrt(n);i++){
        if (!(n%i)) {
            n /= i;
            if (n>i){
                printf("%d\n",n);
                return 0;
            }   
            else{
                printf("%d\n",i);
                return 0;
            }    
        }
    }
}