3529: [Sdoi2014]数表
思路：
莫比乌斯反演+整除分块+树状数组维护前缀和
代码:

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define y1 y11
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
#define u unsigned int
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pli pair<LL, int>
#define pii pair<int, int>
#define piii pair<pii, int>
#define pdd pair<double, double>
#define mem(a, b) memset(a, b, sizeof(a))
#define debug(x) cerr << #x << " = " << x << "\n";
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//head

const int N = 1e5 + 10;
const LL MOD = 1LL<<31;
int T, n, m;
int prime[N], mu[N], g[N], gg[N], cnt;
bool not_p[N];
u ans[N];
struct A {
    int id, g;
    bool operator < (const A & rhs) const {
        return g < rhs.g;
    }
};
struct B {
    int n, m, a, id;
    bool operator < (const B & rhs) const {
        return a < rhs.a;
    }
}q[N];
vector<A> vc;
void seive() {
    mu[1] = 1;
    g[1] = 1;
    for (int i = 2; i < N; ++i) {
        if(!not_p[i]) prime[++cnt] = i, mu[i] = -1, g[i] = 1+i, gg[i] = 1+i;
        for (int j = 1; j <= cnt && i*prime[j] < N; ++j) {
            not_p[i*prime[j]] = true;
            if(i%prime[j] == 0) {
                mu[i*prime[j]] = 0;
                g[i*prime[j]] = g[i]/gg[i];
                gg[i*prime[j]] = gg[i]*prime[j]+1;
                g[i*prime[j]] *= gg[i*prime[j]];
                break;
            }
            mu[i*prime[j]] = -mu[i];
            g[i*prime[j]] = g[i]*(1+prime[j]);
            gg[i*prime[j]] = 1+prime[j];
        }
    }
    for (int i = 1; i < N; ++i) vc.pb(A{i, g[i]});
    sort(vc.begin(), vc.end());
}
struct BIT {
    u bit[N];
    inline void add(int x, u a) {
        while(x < N) bit[x] = bit[x]+a, x += x&-x;
    }
    inline LL sum(int x) {
        LL res = 0;
        while(x) res = res+bit[x], x -= x&-x;
        return res;
    }
}B;
int Q;
int main() {
    seive();
    scanf("%d", &Q);
    for (int i = 1; i <= Q; ++i) {
        scanf("%d %d %d", &q[i].n, &q[i].m, &q[i].a);
        q[i].id = i;
    }
    sort(q+1, q+Q+1);
    int now = 0;
    for (int i = 1; i <= Q; ++i) {
        while(now < vc.size() && vc[now].g <= q[i].a) {
            for (int j = vc[now].id; j < N; j += vc[now].id) {
                B.add(j, vc[now].g*1u*mu[j/vc[now].id]);
            }
            ++now;
        }
        int n = q[i].n, m = q[i].m;
        int up = min(n, m);
        for (int l = 1, r; l <= up; l = r+1) {
            r = min(n/(n/l), m/(m/l));
            u tmp = (n/l)*1u*(m/l)*1u*(B.sum(r)-B.sum(l-1));
            ans[q[i].id] += tmp;
        }
    }
    for (int i = 1; i <= Q; ++i) printf("%lld\n", ans[i]%MOD);
    return 0;
}