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#740772#9619. 乘积,欧拉函数,求和jdyt11TL 265ms4612kbC++202.0kb2024-11-13 11:25:452024-11-13 11:25:45

Judging History

你现在查看的是最新测评结果

  • [2024-11-13 11:25:45]
  • 评测
  • 测评结果:TL
  • 用时:265ms
  • 内存:4612kb
  • [2024-11-13 11:25:45]
  • 提交

answer

#include<bits/stdc++.h>
using namespace std;
#define pii pair<int,int>
#define ll long long
#define inf 0x3f3f3f3f
#define ull unsigned long long
#define pll pair<ll,ll>
#define ls d*2
#define rs d*2+1
#define mid (l+r)/2
#define lowbit(x) (x&(-x))
//#define endl "\n"
#define all(x) x.begin(),x.end()
#define int long long
//mt19937 seed;
//uniform_int_distribution<int>num(0,2e9);
const int N=2e5+10;
const int M=33;

int mod=998244353;
int dp[1<<17];
int pri[20]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};
int qp(int x,int y){
	int re=1;
	while(y){
		if(y&1)re=re*x%mod;
		x=x*x%mod;
		y>>=1;
	}
	return re;
}
signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int _=1;//cin>>_;
    while(_--){
        int n;
        cin>>n;
        map<int,vector<pii>>mp;
        for(int i=0;i<n;i++){
        	int x;
        	cin>>x;
        	int num=0,sum=1;
        	for(int j=0;j<16;j++){
        		while(x%pri[j]==0)num|=(1<<j),sum*=pri[j],x/=pri[j];
			}
			mp[x].push_back({num,sum});
		}
		dp[0]=1;
		for (auto [x, vec] : mp) {
        pri[16] = x;
        for (auto [nd, sum] : vec) {
            if (x != 1)
                nd |= 1 << 16;
            for (int i = (1 << 17) - 1; i >= 0; i--) {
                int nsum = sum, nnd = nd | i;
                if ((i & (1 << 16)) != 0)
                    nsum = (nsum * x) % mod;
                else
                    nsum = (nsum * max(x - 1, 1ll)) % mod;
                dp[nnd] = (dp[nnd] + dp[i] * nsum % mod) % mod;
            }
        }
        for (int i = (1 << 16); i < (1 << 17); i++) {
            int ni = i ^ (1 << 16);
            dp[ni] = (dp[ni] + dp[i]) % mod;
            dp[i] = 0;
        }
    }
		int ans=0;
		for(int i=0;i<(1<<17);i++){
			for(int j=0;j<16;j++){
				if(i>>j&1){
					dp[i]=dp[i]*(pri[j]-1)%mod;
					dp[i]=dp[i]*qp(pri[j],mod-2)%mod;
				}
			}
			ans=(ans+dp[i])%mod;
		}
		cout<<ans<<'\n';
    }
}

详细

Test #1:

score: 100
Accepted
time: 265ms
memory: 4612kb

input:

5
1 6 8 6 2

output:

892

result:

ok single line: '892'

Test #2:

score: 0
Accepted
time: 265ms
memory: 4604kb

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: