LightOJ1109 False Ordering 筛法求约数个数+结构体排序

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 n^th number 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

这题数据量太小啦，所以暴力也能过。

#include <iostream>#include <cstdio>#include <map>#include <set>#include <vector>#include <queue>#include <stack>#include <cmath>#include <algorithm>#include <cstring>#include <string>using namespace std;#define INF 0x3f3f3f3ftypedef long long LL;struct num{    int a;    int b;}s[10005];void p(){    for(int i=1;i<=1000;i++){        s[i].a=i;        s[i].b=1;    }    for(int i=2;i<=1000;i++){        for(int j=i;j<=1000;j+=i){            s[j].b++;        }    }}bool cmp(num x,num y){    return x.b==y.b? x.a>y.a:x.b<y.b;}int main(){std :: ios :: sync_with_stdio (false);    p();    sort(s+1,s+1001,cmp);    int t,n;    scanf("%d",&t);    for(int i=1;i<=t;i++){        scanf("%d",&n);        printf("Case %d: %d\n",i,s[n]);    }    return 0;}