CodeForces - 1512G Short Task(欧拉筛求因子和)

Posted 2021-05-24 Frozen_Guardian

题目链接：点击查看

题目大意：设 $d(n)$ 为 $n$ 的约束和，更具体的，$d(n)={\sum }_{k|n}k$

现在有 $m$ 次询问，每次给出一个 $c$ ，需要回答满足 $d(n)=c$ 的最小 $n$

题目分析：预处理一下就可以了，屯一个 $O(n)$ 求因子和的板子

代码：

// Problem: G. Short Task
// Contest: Codeforces - Codeforces Round #713 (Div. 3)
// URL: https://codeforces.com/contest/1512/problem/G
// Memory Limit: 512 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast","inline","-ffast-math")
// #pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<list>
#include<unordered_map>
#define lowbit(x) x&-x
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
template<typename T>
inline void read(T &x)
{
	T f=1;x=0;
	char ch=getchar();
	while(0==isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(0!=isdigit(ch)) x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
	x*=f;
}
template<typename T>
inline void write(T x)
{
	if(x<0){x=~(x-1);putchar('-');}
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
const int inf=0x3f3f3f3f;
const int N=1e7+100;
int pri[N],psum[N],sum[N],cnt;
bool vis[N];
int ans[N];
void P() {
	sum[1]=1;
	for(int i=2;i<N;i++) {
		if(!vis[i]) {
			pri[cnt++]=i;
			psum[i]=sum[i]=i+1;
		}
		for(int j=0;j<cnt&&pri[j]*i<N;j++) {
			vis[i*pri[j]]=true;
			if(i%pri[j]==0) {
				psum[i*pri[j]]=psum[i]*pri[j]+1;
				sum[i*pri[j]]=sum[i]/psum[i]*psum[i*pri[j]];
				break;
			}
			psum[i*pri[j]]=pri[j]+1;
			sum[i*pri[j]]=sum[i]*psum[i*pri[j]];
		}
	}
}
int main()
{
#ifndef ONLINE_JUDGE
//	freopen("data.in.txt","r",stdin);
//	freopen("data.out.txt","w",stdout);
#endif
//	ios::sync_with_stdio(false);
	P();
	memset(ans,inf,sizeof(ans));
	for(int i=1;i<N;i++) {
		if(sum[i]<N) {
			ans[sum[i]]=min(ans[sum[i]],i);			
		}
	}
	int w;
	cin>>w;
	while(w--) {
		int n;
		read(n);
		printf("%d\\n",ans[n]==inf?-1:ans[n]);
	}
	return 0;
}