感觉这几道数论题都差不多,但这到明显是前几道的升级版。
推了一大顿只能得60分,不得不看题解。
各位看这老哥的题解吧
我就是推到他用\(T\)换掉\(kd\)之前,然后枚举\(T\)的。这个转换确实想不出来啊。
还有最后一句,最终的式子
\[\sum_{T = 1} ^ {n} \lfloor \frac{n}{T} \rfloor \lfloor \frac{m}{T} \rfloor * \sum_{k | T} \mu(\frac{T}{k}) (k \in prime)\]
他把后面的那个sum预处理了。令\(f(T) = \sum_{k | T} \mu(\frac{T}{k}) (k \in prime)\),由此可见,这个函数的自变量是\(T\),而预处理的时候是枚举\(T\)的质因数累加得到\(f(T)\),跟埃氏筛法很像。
#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("")
#define space putchar(‘ ‘)
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 1e7 + 5;
inline ll read()
{
ll ans = 0;
char ch = getchar(), last = ‘ ‘;
while(!isdigit(ch)) last = ch, ch = getchar();
while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - ‘0‘, ch = getchar();
if(last == ‘-‘) ans = -ans;
return ans;
}
inline void write(ll x)
{
if(x < 0) x = -x, putchar(‘-‘);
if(x >= 10) write(x / 10);
putchar(x % 10 + ‘0‘);
}
int v[maxn], prm[maxn], mu[maxn];
ll f[maxn], sum[maxn];
void init()
{
mu[1] = 1;
for(int i = 2; i < maxn; ++i)
{
if(!v[i]) v[i] = i, prm[++prm[0]] = i, mu[i] = -1;
for(int j = 1; j <= prm[0] && i * prm[j] < maxn; ++j)
{
v[i * prm[j]] = prm[j];
if(i % prm[j] == 0) {mu[i * prm[j]] = 0; break;}
else mu[i * prm[j]] = -mu[i];
}
}
for(int i = 1; i <= prm[0]; ++i)
for(int j = 1; prm[i] * j < maxn; ++j)
f[prm[i] * j] += mu[j];
for(int i = 1; i < maxn; ++i) sum[i] = sum[i - 1] + f[i];
}
ll solve(int n, int m)
{
int Min = min(n, m);
ll ret = 0;
for(int l = 1, r; l <= Min; l = r + 1)
{
r = min(n / (n / l), m / (m / l));
ret += (sum[r] - sum[l - 1]) * (n / l) * (m / l);
}
return ret;
}
int main()
{
init();
int T = read();
while(T--)
{
ll n = read(), m = read();
write(solve(n, m)), enter;
}
return 0;
}
原文地址:https://www.cnblogs.com/mrclr/p/10112023.html
时间: 2024-11-07 07:46:45