考虑这样一个式子:
[d(ij) = sum_{x | i}sum_{y | j} [x ot y]
]
怎么证明? 一开始我们一定会想到$$d(ij) = sum_{x | i} sum_{y | j} 1 $$ 但这样会计算重复. 于是我们考虑:
[d(ij) = sum_{x | i}sum_{frac{j}{y} | j}1
]
这样每个因数就变成$ frac{xj}{y} $, 如果x和y不互质. 那么就会有 $ frac{(xj)p}{yp} == frac{xj}{y} $
如果xp,yp同时是i, j的因数那就会算重复.
所以一定要求此二者互质.
相应的:
[sigma(ij) = sum_{x | i}sum_{y | j} [x ot y]x frac{}{} frac{j}{y}
]
那么我们可以开始化式子了!!
[Ans = sum_{i}sum_{j} sum_{x | i}sum_{y | j}[(x, y) == 1] \
= sum_{x} sum_{y}[(x, y) == 1]sum_{i} sum_{j} [x | i][y | j] \
= sum_{x} sum_{y}sum_{d | (x, y)} mu(d) lfloorfrac{n}{x}
floorlfloorfrac{m}{y}
floor\
= sum_{d} mu(d) sum_{x}sum_{y}[d | x][d | y] lfloorfrac{n}{x}
floorlfloorfrac{m}{y}
floor \
= sum_{d} mu(d) sum_{x}^{n / d}lfloorfrac{n}{xd}
floorsum_{y}^{m / d}lfloorfrac{m}{yd}
floor\
]
令(F(n) = sum_{i = 1}^{n} sigma_0(i))
那么有
[Ans = sum_{d} mu(d) F(frac{n}{d})F(frac{m}{d})
]
然后线性筛/杜教筛Min25筛洲阁筛筛以下约数个数的前缀和就可以做了.
Codes
#include<bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = (a), i##_end_ = (b); i <= i##_end_; ++i)
#define drep(i, a, b) for(int i = (a), i##_end_ = (b); i >= i##_end_; --i)
#define clar(a, b) memset((a), (b), sizeof(a))
#define debug(...) fprintf(stderr, __VA_ARGS__)
typedef long long LL;
typedef long double LD;
int read() {
char ch = getchar();
int x = 0, flag = 1;
for (;!isdigit(ch); ch = getchar()) if (ch == '-') flag *= -1;
for (;isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
return x * flag;
}
void write(int x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) write(x / 10);
putchar(x % 10 + 48);
}
const int Maxn = 50009;
int prime[Maxn], isnprime[Maxn], mu[Maxn], prefixMu[Maxn], tot, low[Maxn];
LL sigma0[Maxn];
void linearSieve() {
mu[1] = 1; sigma0[1] = 1;
rep (i, 2, Maxn - 1) {
if (!isnprime[i]) mu[i] = -1, prime[++tot] = i, low[i] = i, sigma0[i] = 2;
for (int k, j = 1; j <= tot && (k = prime[j] * i) < Maxn; ++j) {
isnprime[k] = 1;
if (i % prime[j] == 0) {
mu[k] = 0; low[k] = low[i] * prime[j];
sigma0[k] = (low[i] == i) ? (sigma0[i] + 1) : (sigma0[k / low[k]] * sigma0[low[k]]);
break;
} else {
mu[k] = -mu[i]; low[k] = prime[j];
sigma0[k] = sigma0[i] * sigma0[prime[j]];
}
}
}
rep (i, 1, Maxn - 1) {
prefixMu[i] = prefixMu[i - 1] + mu[i];
sigma0[i] += sigma0[i - 1];
}
}
void init() { linearSieve(); }
void solve() {
int T = read();
while (T--) {
int n = read(), m = read();
LL ans = 0; int Limit = min(n, m);
for (int l = 1, r; l <= Limit; l = r + 1) {
r = min(Limit, min(n / (n / l), m / (m / l)));
ans += (prefixMu[r] - prefixMu[l - 1] * 1ll) * sigma0[n / l] * sigma0[m / l];
}
printf("%lld
", ans);
}
}
int main() {
freopen("BZOJ3994.in", "r", stdin);
freopen("BZOJ3994.out", "w", stdout);
init();
solve();
#ifdef Qrsikno
debug("
Running time: %.3lf(s)
", clock() * 1.0 / CLOCKS_PER_SEC);
#endif
return 0;
}