Graph Theory Class min25筛子

这道题目简化来看就是求3到n+1的所有数和3到n+1的所有质数和，不知道min25筛会很难算，虽然也有打表过去的。

#include<bits/stdc++.h>


using namespace std;
typedef long long ll;

const int mod=998244353;

const int maxn = 1e6 + 10;

bool valid[maxn];
ll prime[maxn];
ll sum[maxn];
int tot;
void init()
{
    
    
    tot = 0;
    valid[1] = true;
    for(int i = 2; i < maxn; i++)
    {
    
    
        if(!valid[i])
        {
    
    
            prime[++tot] = i;
            sum[tot] = sum[tot - 1] + ll(i);
        }
        for(int j = 1; j <= tot && i * prime[j] < maxn; j++)
        {
    
    
            valid[i * prime[j]] = true;
            if(i % prime[j] == 0) break;
        }
    }
}

ll id1[maxn], id2[maxn];
ll w[maxn], g[maxn];

ll solve(ll n, ll md)
{
    
    
    ll sqr = sqrt(n + 0.5);
    //init(sqr);
    int ntot = 0;
    for(ll i = 1; i <= tot; i++)
    {
    
    
        if(prime[i] <= sqr) ntot++;
        else break;
    }
    int m = 0;
    for(ll i = 1; i <= n; i = n / (n / i) + 1)
    {
    
    
        w[++m] = n / i;
        if(w[m] <= sqr) id1[w[m]] = m;
        else id2[n/ w[m]] = m;
        g[m] = w[m] * (w[m] + ll(1));
        g[m] /= ll(2);
        g[m]--;
    }
    for(int j = 1; j <= ntot; j++)
    {
    
    
        for(int i = 1; i <= m && (ll)prime[j] * (ll)prime[j] <= w[i]; i++)
        {
    
    
            ll k = (w[i] / prime[j] <= sqr) ? id1[w[i] / prime[j]] : id2[n / (w[i] / prime[j])];
            g[i] = g[i] - prime[j] * (g[k] - sum[j - 1]);
        }
    }
    return g[id2[1]];
}

int main()
{
    
    
    init();
    int t;
    scanf("%d", &t);
    while(t--)
    {
    
    
        ll n, md;
        scanf("%lld%lld", &n, &md);
        if(n == 1)
        {
    
    
            printf("0\n");
            continue;
        }
        n++;
        ll ans = (n % md) * ((n + 1)% md);
        if(ans & 1) ans += md;
        ans /= ll(2);
        ans %= md;
        ans -= 5;
        ans = (ans + md) % md;
        ans += solve(n, md) % md;
        ans %= md;
        printf("%lld\n", ans);
    }
    return 0;

}
/*
1000
10000000000 998244353
*/