Description
有一张 n×m 的数表，其第 i 行第 j 列（1 <= i <= n, 1 <= j <= m）的数值为
能同时整除 i 和 j 的所有自然数之和。给定 a , 计算数表中不大于 a 的数之和。
Input
输入包含多组数据。
输入的第一行一个整数Q表示测试点内的数据组数
接下来Q行，每行三个整数n，m，a(|a| < =10^9)描述一组数据。
1 < =N．m < =10^5 ， 1 < =Q < =2×10^4
Output
对每组数据，输出一行一个整数，表示答案模2^31的值。
Sample Input
2
4 4 3
10 10 5
Sample Output
20
148
题解%%%%贝神的blog，我就不赘述了
//Author: dream_maker
#include<bits/stdc++.h>
using namespace std;
//----------------------------------------------
//typename
typedef long long ll;
//convenient for
#define fu(a, b, c) for (int a = b; a <= c; ++a)
#define fd(a, b, c) for (int a = b; a >= c; --a)
#define fv(a, b) for (int a = 0; a < (signed)b.size(); ++a)
//inf of different typename
const int INF_of_int = 1e9;
const ll INF_of_ll = 1e18;
//fast read and write
template <typename T>
void Read(T &x) {
bool w = 1;x = 0;
char c = getchar();
while (!isdigit(c) && c != '-') c = getchar();
if (c == '-') w = 0, c = getchar();
while (isdigit(c)) {
x = (x<<1) + (x<<3) + c -'0';
c = getchar();
}
if (!w) x = -x;
}
template <typename T>
void Write(T x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9) Write(x / 10);
putchar(x % 10 + '0');
}
//----------------------------------------------
const int N = 1e5 + 10;
int prime[N], mu[N], tot = 0;
bool vis[N];
int sum[N], pre[N], ans[N];
struct Ques {
int n, m, id, a;
} Q[N];
bool operator < (const Ques a, const Ques b) {
return a.a < b.a;
}
struct Node {
int id, vl;
} P[N];
bool operator < (const Node a, const Node b) {
return a.vl < b.vl;
}
void init() {
mu[1] = sum[1] = pre[1] = 1;
P[1] = (Node){1, 1};
fu(i, 2, N - 1) {
if (!vis[i]) {
prime[++tot] = i;
mu[i] = -1;
sum[i] = pre[i] = i + 1;
}
fu(j, 1, tot) {
int nxt = i * prime[j];
if (nxt >= N) break;
vis[nxt] = 1;
if (i % prime[j]) {
pre[nxt] = prime[j] + 1;
sum[nxt] = sum[i] * sum[prime[j]];
mu[nxt] = -mu[i];
} else {
pre[nxt] = pre[i] * prime[j] + 1;
sum[nxt] = sum[i] / pre[i] * pre[nxt];
mu[nxt] = 0;
break;
}
}
P[i] = (Node){i, sum[i]};
}
}
int bit[N];
void add(int x, int vl) {
for (; x < N; x += x & (-x)) bit[x] += vl;
}
int query(int x) {
int res = 0;
for (; x; x -= x & (-x)) res += bit[x];
return res;
}
int main() {
init();
int T; Read(T);
fu(i, 1, T) {
Read(Q[i].n), Read(Q[i].m), Read(Q[i].a);
Q[i].id = i;
}
sort(Q + 1, Q + T + 1);
sort(P + 1, P + N);
int now = 1;
fu(i, 1, T) {
while (now < N && P[now].vl <= Q[i].a) {
fu(j, 1, (N - 1) / P[now].id) {
add(j * P[now].id, mu[j] * P[now].vl);
}
++now;
}
int limit = min(Q[i].n, Q[i].m), k = 0;
for (int j = 1; j <= limit; j = k + 1) {
k = min(Q[i].n / (Q[i].n / j), Q[i].m / (Q[i].m / j));
ans[Q[i].id] += (Q[i].n / j) * (Q[i].m / j) * (query(k) - query(j - 1));
}
}
fu(i, 1, T) {
Write(ans[i] & 2147483647);
putchar('\n');
}
return 0;
}