Counting Divisors

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题目链接

题意

$d(i)$表示约数函数。计算 .
其中.

思路

因为注意到其实范围不大，我们可以计算这个范围内每一个数字的答案，分开素数和非素数。
素数的话答案就是，非素数我们可以按唯一分解定理将其分成

具体见代码。

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#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#ifndef ONLINE_JUDGE
#define dbg(x...) do{cout << "\033[33;1m" << #x << "->" ; err(x);} while (0)
void err(){cout << "\033[39;0m" << endl;}
template<template<typename...> class T, typename t, typename... A>
void err(T<t> a, A... x){for (auto v: a) cout << v << ' '; err(x...);}
template<typename T, typename... A>
void err(T a, A... x){cout << a << ' '; err(x...);}
#else
#define dbg(...)
#endif
#define inf 1ll << 50
const int N = 1e6 + 100;
ll prime[N];
int tot;
bool vis[N];

void pre()
{
    tot = 0;
    for (int i = 2; i <= 1000010; i++)
    {
        if (!vis[i])
            prime[++tot] = i;
        for (int j = 1; j <= tot && i * prime[j] <= 1000010; j++)
        {
            vis[i * prime[j]] = 1;
            if (i % prime[j] == 0)
                break;
        }
    }
}
ll cnt[N];
ll val[N];
const ll mod = 998244353;
int main()
{
    int T;
    scanf("%d", &T);
    pre();
    while (T--)
    {
        ll l, r, k;
        scanf("%lld%lld%lld", &l, &r, &k);
        for (int i = 0; i <= r - l; i++)
        {
            cnt[i] = 1;
            val[i] = i + l;
        }
        for (int i = 1; i <= tot && prime[i] * prime[i] <= r; i++)
        {
            //dbg(i);
            ll cur = l / prime[i] * prime[i];
            while (cur < l)
            {
                cur = cur + prime[i];
            }
            for (; cur <= r; cur += prime[i])
            {
                ll c = 0;
                while (val[cur - l] % prime[i] == 0)
                {
                    val[cur - l] /= prime[i];
                    c++;
                }
                cnt[cur - l] = (cnt[cur - l] * (c * k % mod + 1)) % mod;
                //dbg(cur, cnt[cur - l + 1]);
            }
        }
        ll ans = 0;
        for (int i = 0; i + l <= r; i++)
        {
            if (val[i] == 1)
                ans = (ans + cnt[i]) % mod;
            else
                ans = (ans + cnt[i] * (k + 1) % mod) % mod;
        }
        printf("%lld\n", ans);
    }
    return 0;
}

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