HDU6069 2017 Multi-University Training Contest - Team 4 1003 Counting Divisors(唯一分解定理+思维优化)

题目大意：

求L到R范围内，每个数K次方的因数个数之和。

思路：

#include<bits/stdc++.h>
using namespace std;
#define pii pair<int, int>
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1000105;
const ll mod = 998244353;
int prim[maxn], p[maxn], cnt = 0;
void init()
{
for(ll i = 2; i <= 1000000; i++)
if(prim[i] == 0)
{
p[++cnt] = i;
for(ll j = i*i; j <= 1000000; j += i)
prim[j] = 1;
}
}
ll number[maxn], val[maxn], tol[maxn];
int main()
{
init();
int t;
ll l, r, k;
scanf("%d", &t);
while(t--)
{
cin >> l >> r >> k;
for(ll i = l; i <= r; i++)
number[i-l+1] = i, val[i-l+1] = 1;
for(ll i = 1; i <= cnt; i++)
{
if(r < p[i]) break;
ll t1 = (l / p[i]) * p[i];
for(; t1 <= r; t1 += p[i])
if(t1 >= l)
{
ll cou = 0;
while(number[t1-l+1] % p[i] == 0)
{
cou++;
number[t1-l+1] /= p[i];
}
val[t1-l+1] = val[t1-l+1] * (cou * k + 1) % mod;
}
}
for(ll i = l; i <= r; i++)
if(number[i-l+1] != 1)
val[i-l+1] = val[i-l+1] * (k + 1) % mod;
ll ans = 0;
for(ll i = l; i <= r; i++)
ans = (ans + val[i-l+1]) % mod;
cout << ans << endl;
}
return 0;
}