【BZOJ3994】【SDOI2015】约数个数和（莫比乌斯反演）

Description

给定nn,mm，求∑i=1n∑j=1md(i×j)∑i=1n∑j=1md(i×j)

其中d(x)d(x)表示xx的约数个数。

Solution

一个结论：

Conclusion：d(nm)=∑i|n∑j|m[gcd(i,j)=1]d(nm)=∑i|n∑j|m[gcd(i,j)=1]

Proof: n=px11px22…n=p1x1p2x2…，m=py11py22…m=p1y1p2y2…，i=pa11pa22…i=p1a1p2a2…，j=pb11pb22…j=p1b1p2b2…

∵gcd(i,j)=1∵gcd(i,j)=1

∴aibi=0∴aibi=0

ai,biai,bi一共有(xi+yi+1)(xi+yi+1)中取值

则总数为∏xi+yi+1∏xi+yi+1

即nmnm的约数和

有了这个结论，就可以开始推式子了：

n∑i=1m∑j=1d(ij)∑i=1n∑j=1md(ij)

=n∑i=1m∑j=1∑x|i∑y|j[gcd(x,y)=1]=∑i=1n∑j=1m∑x|i∑y|j[gcd(x,y)=1]

=n∑i=1m∑j=1⌊ni⌋⌊mj⌋[gcd(i,j)=1]=∑i=1n∑j=1m⌊ni⌋⌊mj⌋[gcd(i,j)=1]

设f(x)=n∑i=1m∑j=1⌊ni⌋⌊mj⌋[gcd(i,j)=x]f(x)=∑i=1n∑j=1m⌊ni⌋⌊mj⌋[gcd(i,j)=x]

F(x)=n∑i=1m∑j=1⌊ni⌋⌊mj⌋[x|gcd(i,j)]=⌊nx⌋∑i=1⌊mx⌋∑j=1⌊nxi⌋⌊mxj⌋F(x)=∑i=1n∑j=1m⌊ni⌋⌊mj⌋[x|gcd(i,j)]=∑i=1⌊nx⌋∑j=1⌊mx⌋⌊nxi⌋⌊mxj⌋

则答案为：f(1)=min{n,m}∑x=1μ(x)⌊nx⌋∑i=1⌊mx⌋∑j=1⌊nxi⌋⌊mxj⌋=min{n,m}∑x=1μ(x)⌊nx⌋∑i=1⌊⌊ni⌋x⌋⌊mx⌋∑j=1⌊⌊mj⌋x⌋f(1)=∑x=1min{n,m}μ(x)∑i=1⌊nx⌋∑j=1⌊mx⌋⌊nxi⌋⌊mxj⌋=∑x=1min{n,m}μ(x)∑i=1⌊nx⌋⌊⌊ni⌋x⌋∑j=1⌊mx⌋⌊⌊mj⌋x⌋

设g(n)=∑ni=1⌊ni⌋g(n)=∑i=1n⌊ni⌋

则f(1)=min{n,m}∑x=1μ(x)g(⌊nx⌋)g(⌊mx⌋)f(1)=∑x=1min{n,m}μ(x)g(⌊nx⌋)g(⌊mx⌋)

对g(i)g(i)预处理或者要求的时候再算然后存下来。

对μ(i)μ(i)线性筛出来然后求前缀和。

最后算的时候整除分块即可。

调了一上午，一直RE，把cout改成printf就A掉了？？

Source

/**************************************
* Au: Hany01
* Date: Feb 2nd, 2018
* Prob: BZOJ3994 & SDOI2015 约数个数和
* Email: hany01@foxmail.com
**************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define fir first
#define sec second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define debug(...) fprintf(stderr, __VA_ARGS__)

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read() {
register int _, __; register char c_;
for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
return _ * __;
}

inline void File() {
#ifdef hany01
freopen("bzoj3994.in", "r", stdin);
freopen("bzoj3994.out", "w", stdout);
#endif
}

const int maxn = 50010;

int n, m, mu[maxn], Sum_mu[maxn], pr[maxn >> 1], cnt, np[maxn];
LL Ans, g[maxn];

inline void Get_mu(int n)
{
mu[1] = 1;
for (register int i = 2; i <= n; ++ i) {
if (!np[i]) pr[++ cnt] = i, mu[i] = -1;
for (register int j = 1; j <= cnt && i * pr[j] <= n; ++ j) {
np[i * pr[j]] = 1;
if (i % pr[j]) mu[i * pr[j]] = -mu[i]; else { mu[i * pr[j]] = 0; break; }
}
}
For(i, 1, n) Sum_mu[i] = Sum_mu[i - 1] + mu[i];
}

inline LL Get_g(int n)
{
if (g
) return g
;
for (register int l = 1, r = 1; l <= n; l = r + 1) r = n / (n / l), g
+= (LL)n / l * (LL)(r - l + 1);
return g
;
}

inline void Solve()
{
for (register int T = read(); T --; ) {
n = read(), m = read(), Ans = 0;
if (n > m) swap(n, m);
for (register int l = 1, r; l <= n; l = r + 1)
r = min(n / (n / l), m / (m / l)), Ans += Get_g(n / l) * Get_g(m / l) * (LL)(Sum_mu[r] - Sum_mu[l - 1]);
printf("%lld\n", Ans);
}
}

int main()
{
File();
Get_mu(maxn - 10);
Solve();
return 0;
}