分析
我们设
为
的所有约数之和
如果不考虑a对答案的影响:
正常的思路,我们枚举
对 提取出个
我们可以使用反演函数替换掉后面那个
对 提取出个
我们令
我们发现 是积性函数,可以用线性筛求出
现在,我们有a的限制:
这样的话,我们只能离线的对所有查询进行处理。首先我们根据a的大小对所以查询排序,a每变大一点,我们都可以插入符合条件的
,这个我们可以用树状数组维护
对于一个符合条件的d,我们可以给d的所有倍数插入相应的值
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
template <typename T>
void out(T x) { cout << x << endl; }
ll fast_pow(ll a, ll b, ll p) {ll c = 1; while(b) { if(b & 1) c = c * a % p; a = a * a % p; b >>= 1;} return c;}
ll exgcd(ll a, ll b, ll &x, ll &y) { if(!b) {x = 1; y = 0; return a; } ll gcd = exgcd(b, a % b, y, x); y-= a / b * x; return gcd; }
const ll N = 1e5 + 100;
const ll mod = 1ll << 31;
struct node
{
ll n, m, a, id;
bool operator < (const node b) const
{
return a < b.a;
}
}q[N], dd[N];
ll prime[N], mu[N], tot;
ll sd[N], sp[N];
bool mark[N];
void get_mu_sd()
{
mark[0] = mark[1] = true;
tot = 0;
mu[1] = sd[1] = sp[1] = 1;
for(ll i = 2; i < N; i ++)
{
if(!mark[i])
{
prime[tot ++] = i;
mu[i] = -1;
sd[i] = sp[i] = i + 1;
}
for(ll j = 0; j < tot && i * prime[j] < N; j ++)
{
mark[i * prime[j]] = true;
if(i % prime[j] == 0)
{
sd[i * prime[j]] = sd[i] / sp[i] * (sp[i] * prime[j] + 1);
sp[i * prime[j]] = sp[i] * prime[j] + 1;
break;
}
sd[i * prime[j]] = (prime[j] + 1) * sd[i];
sp[i * prime[j]] = prime[j] + 1;
mu[i * prime[j]] = -mu[i];
}
}
for(ll i = 1; i < N; i ++)
dd[i].id = i, dd[i].a = sd[i];
sort(dd + 1, dd + N);
}
ll tree[N], ans[N];
ll lowbit(ll x)
{
return x & (-x);
}
void add(ll pos, ll val)
{
for(ll i = pos; i < N; i += lowbit(i))
tree[i] = tree[i] + val;
}
ll query(ll pos)
{
ll num = 0;
for(ll i = pos; i; i -= lowbit(i))
num = num + tree[i];
return num;
}
void add(ll x)
{
for(ll i = 1; i * x < N; i ++)
add(i * x, mu[i] * sd[x]);
}
ll cal(ll n, ll m)
{
ll ans = 0;
if(n > m)
swap(n, m);
for(ll i = 1, j = 1; i <= n; i = j + 1)
{
j = min(n / (n / i), m / (m / i));
ans = (ans + (n / i) * (m / i) % mod * (query(j) - query(i - 1)) % mod + mod) % mod;
}
return ans;
}
int main()
{
ios::sync_with_stdio(false);
get_mu_sd();
memset(tree, 0, sizeof(tree));
ll t;
cin >> t;
for(ll i = 1; i <= t; i ++)
{
cin >> q[i].n >> q[i].m >> q[i].a;
q[i].id = i;
}
sort(q + 1, q + 1 + t);
ll cnt = 1;
for(ll i = 1; i <= t; i ++)
{
while(dd[cnt].a <= q[i].a && cnt < N)
add(dd[cnt ++].id);
ans[q[i].id] = cal(q[i].n, q[i].m);
}
for(ll i = 1; i <= t; i ++)
cout << ans[i] << endl;
}