《洛谷P2522 [HAOI2011]Problem b》

一段时间没做又不太会推了。

首先这题需要一个容斥的做法。

ans = solve(b,d) - solve(a - 1,d) - solve(b,c - 1) + solve(a - 1,c - 1) //这个非常容易推出。

然后就是单个的计算了。

 

$sum_{i = 1}^{n} sum_{j = 1}^{m}|gcd(i,j) = k| ightarrow sum_{i = 1}^{[frac{n}{k}]} sum_{j = 1}^{[frac{m}{j}]}|gcd(i,j) = 1| ightarrow sum_{i = 1}^{[frac{n}{k}]} sum_{j = 1}^{[frac{m}{j}]} sum_{d | gcd(i,j)}^{}mu(d) ightarrow sum_{d = 1}^{min([frac{n}{k}],[frac{m}{j}])}mu (d)~ [frac{n}{kd}]~[frac{m}{kd}]$

看成n / k / d,,m / k / d整除分块即可。

 

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int N = 5e4+5;
const int M = 1e7+5;
const LL Mod = 1e9+7;
#define pi acos(-1)
#define INF 1e9
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline int read()
    {
        int x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-')f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}
        return x * f;
    }
    void print(int x){
        if(x < 0){x = -x;putchar('-');}
        if(x > 9) print(x/10);
        putchar(x%10+'0');
    }
}
using namespace FASTIO;

int a,b,c,d,k;
int prime[N],tot = 0,mu[N];
LL sum[N];
bool vis[N];
void init()
{
    mu[1] = 1;
    for(int i = 2;i < N;++i)
    {
        if(!vis[i])
        {
            prime[++tot] = i;
            mu[i] = -1;
        }
        for(int j = 1;j <= tot && i * prime[j] < N;++j)
        {
            vis[i * prime[j]] = 1;
            if(i % prime[j] == 0) break;
            mu[i * prime[j]] = -mu[i];
        }
    }
    for(int i = 1;i < N;++i) sum[i] = sum[i - 1] + mu[i];
}
LL solve(int n,int m)
{
    n = n / k,m = m / k;
    LL ans = 0;
    for(int L = 1,r = 0;L <= min(n,m);L = r + 1)
    {
        r = min(n / (n / L),m / (m / L));
        ans += (sum[r] - sum[L - 1]) * (n / L) * (m / L);
    }
    return ans;
}
int main()
{
    init();
    int ca;ca = read();
    while(ca--)
    {
        a = read(),b = read(),c = read(),d = read(),k = read();
        LL ans = solve(b,d) - solve(a - 1,d) - solve(b,c - 1) + solve(a - 1,c - 1);
        printf("%lld
",ans);
    }
    system("pause");
    return 0;
}