【LightOJ】1109 - False Ordering（打表）
【LightOJ】1109 - False Ordering（打表）
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1109 - False Ordering
PDF (English) | Statistics | Forum |
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Time Limit: 1 second(s) | Memory Limit: 32 MB |
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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 | 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 |
打一个因子表，sort 后就好了。
代码如下：
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(a,b) memset(a,b,sizeof(a))
#define LL long long
#define PI acos(-1.0)
struct node
{
int n,m;
}a[1005];
bool cmp(node x,node y)
{
if (x.m == y.m)
return x.n > y.n;
return x.m < y.m;
}
int main()
{
int u;
int n;
int Case = 1;
for (int i = 1 ; i <= 1000 ; i++)
{
a[i].n = i;
a[i].m = 1;
}
for (int i = 2 ; i <= 1000 ; i++)
{
for (int j = i ; j <= 1000 ; j += i)
a[j].m++;
}
sort(a+1,a+1001,cmp);
scanf ("%d",&u);
while (u--)
{
scanf ("%d",&n);
printf ("Case %d: ",Case++);
printf ("%d\n",a[n].n);
}
return 0;
}腾讯云开发者
