关于此题Codeforces Round 1003 (Div. 4)_G. Skibidus and Capping

记录解决这道题过程中先后的疏忽

#include<bits/stdc++.h>
    
using namespace std;
    
long long t;
const long long N = 2e5 + 10;
long long n,a[N],ans;
long long Prime[N],tot,pos[N],p[N];
long long sum[N],num,h[N];
bool vis[N],v[N];

void work() {
    for(long long i = 2;i < N;i++) {
        if(!vis[i]) Prime[++tot] = i;
        for(long long j = 1;j <= tot && i * Prime[j] < N;j++) {
            vis[i * Prime[j]] = 1;
            if(i % Prime[j] == 0) break;
        }
    }
}
    
void solve() {
    for(long long i = 0;i <= n;i++) pos[i] = ans = sum[i] = num = v[i] = 0;
    cin >> n;
    for(long long i = 1;i <= n;i++) cin >> a[i];
    long long cnt = 0;
    for(long long i = 1;i <= n;i++) 
        if(!vis[a[i]]) cnt++,p[a[i]]++;
        else h[a[i]]++;                                             //漏了相同合数之间可以作为一对
    for(long long i = 1;i <= n;i++)
        if(!vis[a[i]] && p[a[i]]) sum[++num] = p[a[i]],p[a[i]] = 0;
    for(long long i = 1;i <= num;i++) {
        cnt -= sum[i];                                              //质数之间统计答案出错
        ans += cnt * sum[i];
    }
    for(long long i = 1;i <= n;i++)
        if(vis[a[i]])
            for(long long j = 2;j <= sqrt(a[i]);j++) {
                if(a[i] % j == 0) {
                    long long tmp = a[i] / j;
                    if(!vis[tmp] && !vis[j]) {
                        pos[j]++;
                        if(j != tmp) pos[tmp]++;
                        ans++;                                                  //漏了合数自己跟自己可以构成一对答案
                        v[a[i]] = 1;
                        break;
                    }
                }
            }
    for(int i = 1;i <= n;i++)
        if(h[a[i]] && v[a[i]]) ans += h[a[i]] * (h[a[i]] - 1) / 2,h[a[i]] = 0;  //合数本身必须满足题意才能在相同合数之间配对
    for(long long i = 1;i <= n;i++)
        if(!vis[a[i]]) ans += pos[a[i]];
    cout << ans << '\n';
}
    
signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    work();
    cin >> t;
    while(t--) solve();
    
    return 0;
}