LightOJ 1109 - False Ordering (因子数打表排列)

1109 - False Ordering

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Time Limit: 1 second(s)

Memory Limit: 32 MB

We define b is a Divisor of a number a if
ais divisible by b. So, the divisors of 12are 1, 2, 3, 4, 6, 12. So, 12 has 6 divisors.

Now you have to order all theintegers from 1 to 1000. x will come before y if

1)                 numberof divisors of x is less than number of divisors ofy

2)                 numberof divisors of x is equal to number of divisors ofy and
x> y.

Input

Input starts with an integer T (≤ 1005),denoting the number of test cases.

Each case contains an integer n (1 ≤ n ≤1000).

Output

For each case, print the case number and the nthnumber after ordering.

Sample Input	Output for Sample Input
5 1 2 3 4 1000	Case 1: 1 Case 2: 997 Case 3: 991 Case 4: 983 Case 5: 840

 

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010100
#define LL long long
#define ll __int64
#define INF 0x7fffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-10
using namespace std;
LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
struct s
{
int num;
int k;
}ans[1010];
bool cmp(s a,s b)
{
if(a.k==b.k)
return a.num>b.num;
return a.k<b.k;
}
void db()
{
int i,j;
for(i=1;i<=1000;i++)
{
ans[i].num=i;
ans[i].k=1;
int x=i;
int n=i;
for(j=2;j<=sqrt(n);j++)
{
if(x%j==0)
{
int cnt=0;
while(x%j==0)
{
cnt++;
x/=j;
}
ans[i].k=ans[i].k*(cnt+1);
}
}
if(x>1)
ans[i].k*=2;
}
sort(ans,ans+1001,cmp);
}
int main()
{
db();
int t,i,j;
int n,k;
int cas=0;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
printf("Case %d: ",++cas);
printf("%d\n",ans
.num);
}
return 0;
}