周赛

H 圆周率用acos(-1.0) 使用longlong
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu
Submit Status Practice LightOJ
1109 

uDebug

Description

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

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

1)                  number of divisors of x is less than number of divisors of y

2)                  number of divisors of x is equal to number of divisors of y 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 nth number after ordering.

Sample Input

5

1

2

3

4

1000

Sample Output

Case 1: 1

Case 2: 997

Case 3: 991

Case 4: 983

Case 5: 840

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
struct node{
int shi,jia;
}shu[1010];

void geshu()
{
for(int i=1;i<=1000;i++)
{
int sum=0;
for(int j=1;j<=i;j++)
{
if(i%j==0)
sum++;
}
shu[i].jia=sum;
shu[i].shi=i;
}
}
bool cmp(node x,node y)
{
if(x.jia==y.jia){//这里要逻辑清晰；；；；
return x.shi>y.shi;
}
else{
return x.jia<y.jia;
}
}

int main()
{
int t,n,cont=0;
scanf("%d",&t);
geshu();
sort(shu+1,shu+1000+1,cmp);
while(t--)
{
scanf("%d",&n);
printf("Case %d: %d\n",++cont,shu
.shi);
}
return 0;
}//HHHHHHHHHH