URAL 2047 Maths 打表 递推

Maths
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87157#problem/B

Description

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, …, a_n such that for any k from 2 to n the sum a₁ + … + a_k has exactly a_k different positive divisors. Help Vasya to cope with this task.

Input

The only line contains an integer n (2 ≤ n ≤ 100 000).

Output

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

Sample Input

3

Sample Output

1 3 4

HINT

 

题意

让你构造n个数，要求a1+……+ak的和的因子数恰好为ak

题解：

假设我们已经构造出了ak,且前缀和sum已知，那么ak-1=sum-ak，这个是显然的结论

于是我们直接打表打出来100000位就好了，然后把sum求出来，然后前面直接递推就好了

代码：

打表程序：

#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)  
#define maxn 2000001
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}
//**************************************************************************************int cnt[300*200010];
vector<int> Q;
int flag=0;
int n;
void dfs(int N,int sum,int z)
{if(flag)return;if(sum+z>300*100000)return;if(N>1&&cnt[sum]!=z)return;if(N==n){printf("%d",Q[0]);for(int i=1;i<Q.size();i++)printf(" %d",Q[i]);printf("\n");cout<<sum<<endl;flag=1;return;}for(int i=1;i<=300;i++){Q.push_back(i);dfs(N+1,sum+i,i);Q.erase(Q.end()-1);}}
int main()
{freopen("1.out","w",stdout);for(int i=1;i<=300*100010;i++){for(int j=i;j<300*100010;j+=i){cnt[j]++;}}n=read();Q.clear();flag=0;for(int i=1;i<=300;i++){Q.push_back(i);dfs(1,i,i);Q.erase(Q.end()-1);}if(flag==0)printf("Impossible\n");
}

AC程序：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)  
#define maxn 20001
#define mod 1000000007
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{ll x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}
//**************************************************************************************
const int N=2000000;
int cnt[N];
int a[N],sum=0;
vector<int> Q;
int flag=0;
int n;
int tot;
int main()
{for(int i=1;i<N;i++){for(int j=i;j<N;j+=i)cnt[j]++;}for(int p=1586335,i=100000;i>0;p-=cnt[p],i--){a[i]=cnt[p];sum+=a[i];}n=read();for(int i=1;i<n;i++) printf("%d ",a[i]);printf("%d\n",a[n]);
}