HDU6069 2017 Multi-University Training Contest

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：

唯一分解定理

不要对L-R之间内的数一次次地进行分解质因数，要把质因数的循环放外面，去试除L-R的每个数，这样才能做到一大波常数优化。

#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;}