[POI2007]ZAP-Queries && [HAOI2011]Problem b 莫比乌斯反演

1，[POI2007]ZAP-Queries

～～～题面～～～
题解：　首先列出式子：$$ans = sum_{i = 1}^{n}sum_{j = 1}^{m}[gcd(i, j) == d]$$
　　　　$$[gcd(i, j) == d] = [gcd(lfloor{frac{i}{d}} floor,lfloor{frac{j}{d}} floor) == 1]$$
　　　　所以原式 $$Rightarrow quad sum_{i = 1}^{lfloor{frac{n}{d}} floor}sum_{j = 1}^{lfloor{frac{m}{d}} floor}[gcd(i, j)==1]$$
　　　　$$Rightarrow quad sum_{i = 1}^{n}sum_{j = 1}^{m}sum_{d|gcd(i, j)}mu(d)$$
　　　　因为$mu(d)$会被统计到当且仅当$d | i quad and quad d | j$,即$d | gcd(i, j)$
　　　　那么考虑将满足条件的i和j两两搭配，组成的方案数就是$mu(d)$会被统计到的次数，
　　　　也就是$mu(d)$会被统计到$lfloor{frac{n}{d}} floorlfloor{frac{m}{d}} floor$次
　　　　$$Rightarrow quad ans=sum_{i=1}^{min(n,m)}{mu(d)lfloor{frac{n}{d}} floorlfloor{frac{m}{d}} floor}$$
　　　　然后观察到$lfloor{frac{n}{d}} floorlfloor{frac{m}{d}} floor$中有很多小段$lfloor{frac{n}{d}} floorlfloor{frac{m}{d}} floor$乘积是固定的，也就是$lfloor{frac{n}{d}} floor$和$lfloor{frac{m}{d}} floor$同时为一个固定的值，因此我们可以用整数分块优化

#include<bits/stdc++.h>
using namespace std;
#define R register int
#define AC 55000
int t, n, m, d;
int prime[AC], mu[AC], sum[AC], tot;
bool z[AC];

inline int read()
{
    int x = 0;char c = getchar();
    while(c > '9' || c < '0') c = getchar();
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x;
}

void getprime()
{
    int now;
    mu[1] = 1;
    for(R i = 2; i <= 50000; i++)
    {
        if(!z[i]) prime[++tot] = i, mu[i] = -1;
        for(R j = 1; j <= tot; j++)
        {
            now = prime[j];
            if(now * i > 50000) break;
        //    printf("!!!%d %d
", now, i);
            z[now * i] = true;
            if(!(i % now)) break;
            mu[i * now] = -mu[i];
        }
    }
    for(R i = 1; i <= 50000; i++) 
        sum[i] = mu[i] + sum[i - 1];
}

int ans;

void work()
{
    int pos;
    t = read();
    for(R i = 1; i <= t; i++)
    {
        n = read(), m = read(), d = read();
        n /= d, m /= d;
        int b = min(n, m);
        ans = 0;
        for(R j = 1; j <= b; j = pos + 1)
        {
            pos = min(n / (n / j), m / (m / j));
            ans += (sum[pos] - sum[j-1]) * (n / j) * (m / j);            
        }   
        printf("%d
", ans);
    }
}

int main()
{
//    freopen("in.in", "r", stdin);
    getprime();
    work();
//    fclose(stdin);
    return 0;
}

2,[HAOI2011]Problem b

～～～题面～～～
题解：　这题相比上题多出了两个限制条件，同时对上下限都有限制，那么此时可以用一个容斥来求。
　　　　设$cal(n,m)$表示满足$x le n$和$y le m$且$gcd(x, y) = d$的点对个数，
　　　　那么$$ans = cal(b, d) - cal(a - 1, d) - cal(b, c - 1) + cal(a - 1, c - 1);$$

#include<bits/stdc++.h>
using namespace std;
#define R register int
#define AC 55000
#define LL long long
int a, b, c, d, tot, k, t, ans;
int prime[AC], mu[AC], sum[AC];
bool z[AC];

inline int read()
{
    int x = 0; char c = getchar();
    while(c > '9' || c < '0') c = getchar();
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x;
}

void pre()
{
    int now;
    mu[1] = 1;
    for(R i = 2; i <= 50000; i++)
    {
        if(!z[i]) prime[++tot] = i, mu[i] = -1;//质数的mu值肯定都是-1
        for(R j = 1; j <= tot; j++)
        {
            now = prime[j];
            if(now * i > 50000) break;
            z[now * i] = true;
            if(!(i % now)) break;
            mu[i * now] = -mu[i];//error这里的mu是由mu[i]得来的啊！！！
        }
    }
    for(R i = 1; i <= 50000; i++) sum[i] = sum[i - 1] + mu[i];
}

int cal(int n, int m)
{
    n /= k, m /= k;
    int b = min(n, m), pos, rnt = 0;
    for(R i = 1; i <= b; i = pos + 1)
    {
        pos = min(n / (n / i), m / (m / i));
        rnt += (sum[pos] - sum[i-1]) * (n / i) * (m / i);
    }
    return rnt;
}
//ans = cal(b, d) - cal(a - 1, d) - cal(b, c - 1) + cal(a - 1 , c - 1)，相当于容斥 

void work()
{
    t = read();
    for(R i = 1; i <= t; i++)
    {
        a = read(), b = read(), c = read(), d = read(), k = read();
        ans = cal(b, d) - cal(a - 1, d) - cal(b, c - 1) + cal(a - 1, c - 1);
        printf("%d
", ans);
    }
}

int main()
{
//    freopen("in.in", "r", stdin);
    pre();
    work();
//    fclose(stdin);
    return 0;
}