湖大训练赛8 Identification Codes

Identification CodesTime Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KBTotal submit users: 6, Accepted users: 6Problem 12892 : No special judgementProblem description

MI6 uses a Spy Identification Code (SIC) to identify their spies. For example, J. B.2 has a SIC of 7. The SICs have been assigned to the spies in such a way that MI6 can easily refer to any group of spies by using a status code that is the product of all SICs of the spies in the group.
More precisely, the SICs are chosen in such a way that each status code  2 refers to a unique group of spies, and for each group of spies there is a unique status code referring to it. Write a program that, given a status code, returns the SICs of the spies that belong to the group corresponding to that status code.

Input

On the first line one positive number: the number of test cases, at most 100. After that per test case:
 one line with an integer c (2<=c<=10^9): the status code.

Output

one line with a space-separated list of SICs for the status code, in increasing order.

Sample Input

571264721337

Sample Output

73 44 162 4 97 191

唉。一个看样例猜题意的题目。题意我也不是怎么说得清楚。大致意思的是每个间谍都有自己的部门。啊，不说了。说说解法算了。看样例知道都是n这个数的因子。然后基本都是素数或者倍数。比如k的平方倍，为什么又没有k的立方倍呢？(详情请看样例3)2和4可以组成8也就是2的立方倍，显然，可以大胆的假设是平方倍的平方倍才可以。或者素数，这个当然大家都知道。于是代码就可以敲出来了。用素数筛选法把10e5前的素数晒出来，然后对n进行判断是否是因子，然后看是否有倍数的倍数存在。然后就额米有然后了。RE了一两次的原因还是再算倍数和倍数的时候h定义成了int类型太小了，导致了负数的产生。结果就悲剧了，改成__int64就可以Accept了。看到Accept还是有点小小的激动。

#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <iostream>#include <cmath>#include <queue>#include <map>#include <stack>#include <list>#include <vector>using namespace std;#define LL __int64int t;int a[100020],p[100000],b[100000];void sf(long long  n){memset(a,0,sizeof(a));memset(p,0,sizeof(p));t=0;for (int i=2;i<n;i++){if (!a[i])p[t++]=i;for (int j=0;p[j] && (p[j]*i)<=n;j++){a[p[j]*i]=1;if (i % p[j]==0) break;}}} int main(){sf(100000);int T,i;LL n;scanf("%d",&T);while (T--){scanf("%I64d",&n);int k=0,ff=0;for (i=0;i<t;i++){if (n==1) {ff=1;break;}if (n <= p[i]){ff=1;b[k]=n;k++;break;}else{if (n % p[i]!=0) continue;else{while (n % p[i]==0){LL h=p[i];while (n%h==0){b[k]=h;h*=h;}n/=sqrt(h);k++;}}}}if (ff==0) b[k++]=n;sort(b,b+k);for (i=0;i<k-1;i++)printf("%d ",b[i]);cout<<b[k-1]<<endl;}return 0;}