P2257 YY的GCD

Posted 2021-09-08 Jozky86

P2257 YY的GCD

题意：

求 $1\le x\le N,1\le y\le M$ 且gcd(x, y) 为质数的 (x,y) 有多少对。

题解：

莫比乌斯反演

代码：

#include <bits/stdc++.h>
#include <unordered_map>
#define debug(a, b) printf("%s = %d\\n", a, b);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
clock_t startTime, endTime;
//Fe~Jozky
const ll INF_ll= 1e18;
const int INF_int= 0x3f3f3f3f;
void read(){};
template <typename _Tp, typename... _Tps> void read(_Tp& x, _Tps&... Ar)
{
    x= 0;
    char c= getchar();
    bool flag= 0;
    while (c < '0' || c > '9')
        flag|= (c == '-'), c= getchar();
    while (c >= '0' && c <= '9')
        x= (x << 3) + (x << 1) + (c ^ 48), c= getchar();
    if (flag)
        x= -x;
    read(Ar...);
}
template <typename T> inline void write(T x)
{
    if (x < 0) {
        x= ~(x - 1);
        putchar('-');
    }
    if (x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
void rd_test()
{
#ifdef LOCAL
    startTime= clock();
    freopen("in.txt", "r", stdin);
#endif
}
void Time_test()
{
#ifdef LOCAL
    endTime= clock();
    printf("\\nRun Time:%lfs\\n", (double)(endTime - startTime) / CLOCKS_PER_SEC);
#endif
}
const int maxn= 1e7 + 9;
ll mu[maxn];
int vis[maxn], prim[maxn], cnt, f[maxn];
int sum[maxn];
void get_mu(int n)
{
    mu[1]= 1;
    for (int i= 2; i <= n; i++) {
        if (!vis[i]) {
            prim[++cnt]= i;
            mu[i]= -1;
        }
        for (int j= 1; j <= cnt && prim[j] * i <= n; j++) {
            vis[prim[j] * i]= 1;
            if (i % prim[j] == 0)
                break;
            else
                mu[i * prim[j]]= -mu[i];
        }
    }
    for (int i= 1; i <= cnt; i++) {
        for (int j= 1; prim[i] * j <= n; j++) {
            f[j * prim[i]]+= mu[j];
        }
    }
    for (int i= 1; i <= n; i++)
        sum[i]= sum[i - 1] + f[i];
}
ll solve(int a, int b)
{
    ll ans= 0;
    for (int l= 1, r= 0; l <= a; l= r + 1) {
        r= min(a / (a / l), b / (b / l));
        ans+= 1ll * (sum[r] - sum[l - 1]) * 1ll * (a / l) * (b / l);
    }
    return ans;
}
int main()
{
    //rd_test();
    get_mu(10000000);
    int t;
    read(t);
    while (t--) {
        int n, m;
        read(n, m);
        if (n > m)
            swap(n, m);
        printf("%lld\\n", solve(n, m));
    }
    //Time_test();
}