P3455 [POI2007]ZAP-Queries

对于这种与gcd有关的莫比乌斯反演，一般我们都是套路的去设f(d)为gcd(i,j)=d的个数，F(n)为gcd(i,j)=d和d的倍数的个数,然后用莫比乌斯反演,然后整出一个可以整数分块的东西.

等我学完latex再发式子,详情可见luogu题解.

题干:

FGD正在破解一段密码，他需要回答很多类似的问题：对于给定的整数a,b和d，有多少正整数对x,y，满足x<=a，y<=b，并且gcd(x,y)=d。作为FGD的同学，FGD希望得到你的帮助。

代码:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
#include<algorithm>
#include<vector>
#include<complex>
#include<cstring>
using namespace std;
#define duke(i,a,n) for(int i = a;i <= n;i++)
#define lv(i,a,n) for(int i = a;i >= n;i--)
#define clean(a) memset(a,0,sizeof(a))
#define mp make_pair
#define cp complex<db>
#define enter puts("")
const long long INF = 1LL << 60;
const double eps = 1e-8;
typedef long long ll;
typedef double db;
template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-') op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10 + c - '0';
    if(op) x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar('0' + x % 10);
}
const int N = 50005;
int miu[N],sum[N],tot = 0,che[N],pri[N];
void init()
{
    miu[1] = 1;
    duke(i,2,N)
    {
        if(!che[i])
        {
            pri[++tot] = i;
            miu[i] = -1;
        }
        duke(j,1,tot)
        {
            if(i * pri[j] > N) break;
            che[i * pri[j]] = 1;
            if(!(i % pri[j]))
            break;
            else
            miu[pri[j] * i] = -miu[i];
        }
    }
    duke(i,1,N)
    sum[i] = sum[i - 1] + miu[i];
}
int T;
int main()
{
    read(T);
    init();
    while(T--)
    {
        ll n,m,d,ans = 0;
        read(n);read(m);read(d);
        n /= d;m /= d;
        if(n < m) swap(n,m);
        for(int i = 1;i <= m;)
        {
            int t;
            t = min(n,min(n / (n / i),m / (m / i)));
            ans += (sum[t] - sum[i - 1]) * (n / i) * (m / i);
            i = t + 1;
        }
        printf("%lld
",ans);
    }
    return 0;
}