P1447 [NOI2010]能量采集
式子化简
显然题目就是要我们求
转化为我们要求
套上
另
反演有
代码
/*
Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;
inline ll read() {
ll f = 1, x = 0;
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 f * x;
}
const int N = 1e7 + 10;
bool st[N];
vector<int> prime;
int n, m;
ll phi[N];
void mobius() {
st[0] = st[1] = phi[1] = 1;
for(int i = 2; i < N; i++) {
if(!st[i]) {
prime.pb(i);
phi[i] = i - 1;
}
for(int j = 0; j < prime.size() && i * prime[j] < N; j++) {
st[i * prime[j]] = 1;
if(i % prime[j] == 0) {
phi[i * prime[j]] = phi[i] * prime[j];
break;
}
phi[i * prime[j]] = phi[i] * (prime[j] - 1);
}
}
for(int i = 1; i < N; i++) phi[i] += phi[i - 1];
}
int main() {
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
mobius();
ll n = read(), m = read();
if(n > m) swap(n, m);
ll ans = 0;
for(ll l = 1, r; l <= n; l = r + 1) {
r = min(n / (n / l), m / (m / l));
ans += (n / l) * (m / l) * (phi[r] - phi[l - 1]);
}
printf("%lld\n", 2 * ans - n * m);
return 0;
}