URAL2047 Maths (暴力打表 递推)

2047. Maths

Time limit: 1.0 second
Memory limit: 64 MB

Android Vasya attends Maths classes. His group started to study the number theory recently. The teacher gave them several tasks as a homework. One of them is as follows.
There is an integer n. The problem is to find a sequence of integers a1,…,an such that for any k from 2 to n the sum a1+…+ak has exactly ak different positive divisors. Help Vasya to cope with this task.

Input

The only line contains an integer n(2≤n≤100000).

Output

If there is no such sequence output “Impossible”. Otherwise output space-separated integers a1,…,an(1≤ai≤300).

Sample

input

3

output

1 3 4

题意

输入一个n,构造一个长度为n的序列，要求是对于任意一个ak(2≤k≤n),a1+a2+......+ak有ak个因数(a≤300)。

题解

首先暴力打个表，看看按照要求能够造出的最长的序列是多少，发现可以构造出长度为100000的序列，记下满足条件的最小的前缀和，然后打表逆推就行了，对于任意一个ak,ak−1=sumk−num[sumk],num[sumk]是sumk的因数个数。(被爆栈折磨了一晚上。。。)
打表代码:

#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <stack>using namespace std;int num[30000005];int MAX = 100000000;int dfs(int n, int sum){    if (n == 100000)    {        MAX = min(MAX, sum);        return 1;    }    for (int i = 1; i <= 300; i++)    {        if (sum + i>30000005)            return 0;        if (num[sum + i] == i&&dfs(n + 1, sum + i))            return 1;    }    return 0;}int main(){    memset(num, 0, sizeof num);    for (int i = 1; i <= 30000005; i++)    {        for (int j = i; j <= 30000005; j += i)            num[j]++;    }    for (int i = 1; i <= 300; i++)    {        dfs(1, i);    }    printf("%d\n", MAX);    getchar();    return 0;}

AC代码:

#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <stack>using namespace std;int num[1586336];int ans[100005];int n;int main(){    memset(num,0,sizeof num);    for(int i=1;i<=1586335;i++)    {        for(int j=i;j<=1586335;j+=i)            num[j]++;    }    int sum=1586335;    for(int i=100000;i>0;i--)    {        ans[i]=num[sum];        sum-=num[sum];    }    scanf("%d",&n);    for(int i=1;i<=n;i++)        printf("%d\n",ans[i]);    return 0;}