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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#810928#9619. 乘积,欧拉函数,求和Nana7TL 111ms6292kbC++142.1kb2024-12-12 13:32:572024-12-12 13:32:58

Judging History

你现在查看的是最新测评结果

  • [2024-12-12 13:32:58]
  • 评测
  • 测评结果:TL
  • 用时:111ms
  • 内存:6292kb
  • [2024-12-12 13:32:57]
  • 提交

answer

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#define I inline
#define int long long
using namespace std;

const int mod = 998244353;
const int MAXN = 65600;
const int V = 3010;
const int N = 2010;
int a[N],n,pri[N],cnt,f[2][MAXN],mask[V],mul[MAXN],g[2][MAXN];
int pw[22],id[N];
vector<int> v[V];

int qpow(int x,int y) {
	if(y==0) return 1;
	if(y==1) return x;
	int mk=qpow(x,y/2);
	return y%2?mk*mk%mod*x%mod:mk*mk%mod;
}
/*
3
1 2 3

*/
I void solve() {
	cin>>n;
	for(int i=1;i<=n;++i) cin>>a[i];
	int lst=0; f[0][0]=1;
	for(int i=1;i<=n;++i) {
		
		int now=lst^1,mk=a[i];
		for(int j=1;j<=16;++j) {
			while(mk%pri[j]==0) mk/=pri[j];
		}
		
		if(mk!=1) {
			v[mk].push_back(a[i]);
			continue;
		}
		
		for(int j=0;j<pw[16];++j) f[now][j]=f[lst][j];
		for(int j=0;j<pw[16];++j) {
			int val=mul[mask[a[i]]-(j&mask[a[i]])];
			f[now][j|mask[a[i]]]+=f[lst][j]*val%mod*a[i]%mod;
		}
		lst^=1;
	}
	
	for(int i=1;i<=3000;++i) {
		
		if(!v[i].size()) continue;
		
		for(int j=0;j<pw[16];++j) {
			g[0][j]=f[lst][j];
			g[1][j]=0;
		}
		
		for(auto &t:v[i]) {
			//int fac=i,sfac=t/i;
			for(int j=pw[16];j>=0;--j) {
				int val=mul[mask[t/i]-(j&mask[t/i])];
				g[1][j|mask[t/i]]+=g[1][j]*t%mod*val%mod;
				g[1][j|mask[t/i]]+=g[0][j]*t%mod*qpow(i,mod-2)%mod*(i-1)%mod*val%mod;
				g[1][j|mask[t/i]]%=mod;
			}
		}
		for(int j=0;j<pw[16];++j) {
			f[lst][j]=g[0][j]+g[1][j];
			f[lst][j]%=mod;
		}
	}
	int ans=0;
	for(int i=0;i<pw[16];++i) {
		ans+=f[lst][i];
		ans%=mod;
	}
	cout<<ans<<endl;
}
I void init() {
	for(int i=2;i*i<=3000;++i) {
		bool can=1;
		for(int j=2;j*j<=i;++j) {
			if(i%j==0) can=0;
		}
		if(can) pri[++cnt]=i,id[i]=cnt;
	}
	pw[0]=1;
	for(int i=1;i<=16;++i) pw[i]=pw[i-1]<<1;
	for(int i=1;i<=3000;++i) {
		for(int j=1;j<=16;++j) {
			if(i%pri[j]==0) mask[i]|=pw[j-1];
		} 
	}
	for(int i=0;i<pw[16];++i) {
		mul[i]=1;
		for(int j=1;j<=16;++j) {
			if(i&pw[j-1]) {
				mul[i]*=qpow(pri[j],mod-2)*(pri[j]-1)%mod;
				mul[i]%=mod;
			}
		}
	}
}

signed main() 
{
	init();
	solve();
} 

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 107ms
memory: 5316kb

input:

5
1 6 8 6 2

output:

892

result:

ok single line: '892'

Test #2:

score: 0
Accepted
time: 111ms
memory: 6292kb

input:

5
3 8 3 7 8

output:

3157

result:

ok single line: '3157'

Test #3:

score: -100
Time Limit Exceeded

input:

2000
79 1 1 1 1 1 1 2803 1 1 1 1 1 1 1609 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2137 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 613 1 499 1 211 1 2927 1 1 1327 1 1 1123 1 907 1 2543 1 1 1 311 2683 1 1 1 1 2963 1 1 1 641 761 1 1 1 1 1 1 1 1 1 1 1 1489 2857 1 1 1 1 1 1 1 1 1 1 1 1 1 967 1 821 1 1 1 1 2143 1861...

output:


result: