HDU 5108 最大素數因子

Alexandra and Prime Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 706 Accepted Submission(s): 247


Problem Description Alexandra has a little brother. He is new to programming. One day he is solving the following problem: Given an positive integer N, judge whether N is prime.
The problem above is quite easy, so Alexandra gave him a new task: Given a positive integer N, find the minimal positive integer M, such that N/M is prime. If such M doesn't exist, output 0.
Help him!
Input There are multiple test cases (no more than 1,000). Each case contains only one positive integer N.
N≤1,000,000,000.
Number of cases with N>1,000,000 is no more than 100.
Output For each case, output the requested M, or output 0 if no solution exists.
Sample Input

3
4
5
6

Sample Output

1
2
1
2

Source BestCoder Round #19
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題意 ： 給你一個N 尋找一個最小M ，使得N/M=素數 不解釋了，簡單代碼 #include
#include
#include
#include
using namespace std;
int f(int x)
{
int flag = 0;
for(int i = 2; (i*i) <= x; i ++)
{
if(x % i == 0)
{
flag = 1;
break;
}
}
if(flag) return 0;
return 1;
}
int main()
{
int n;
int mark[200005];
while(scanf("%d",&n)!=EOF)
{
if( n == 1 ) {printf("0\n"); continue;}
int count = 0, flag = 0;
for(int i = 1; (i * i) <= n; i++)
{
if( n%i == 0) {mark[count++] = i; mark[count++] = n / i;}
}
sort(mark, mark + count);
int i;
for(i = count - 1 ; i >= 0; i --)
{
if(f(mark[i])) { break;}
}
printf("%d\n",n / mark[i]);
}
}