题目链接
思路
如果这题是这样的：
[F(n)=sumlimits_{i=1}^{n}sumlimits_{j=1}^{n}phi(gcd(i,j))
]
那么我们可能会想到下面方法进行反演：
[egin{aligned}
F(n)=&sumlimits_{k=1}^{n}phi(k)sumlimits_{i=1}^{n}sumlimits_{j=1}^{n}[gcd(i,j)=k]&\
=&sumlimits_{k=1}^{n}phi(k)sumlimits_{i=1}^{frac{n}{k}}sumlimits_{j=1}^{frac{n}{k}}[gcd(i,j)=1]&\
end{aligned}
]
令(f(n))为(gcd(i,j)=n)的有序对的对数，(g(n))为(gcd(i,j)=n ext{和}n)的倍数的有序对的对数，则
[egin{aligned}
&g(n)=sumlimits_{n|d}f(d)&\
ightarrow&f(n)=sumlimits_{n|d}mu(frac{d}{n})g(d)&
end{aligned}
]
然后将(f(1)=sumlimits_{i=1}^{n}mu(i)g(i))代入(F(n))得到(F(n)=sumlimits_{k=1}^{n}phi(k)sumlimits_{i=1}^{frac{n}{k}}mu(i)g(i))，然后先预处理(sumlimits_{i=1}^{frac{n}{k}}mu(i)g(i))，然后暴力枚举(k)或者数论分块求解答案就可以了，这里有几道类似的题目，有兴趣的可以做一下。
但是这题却不是我们所希望的那种形式，那么该怎么办呢？我们发现(n)小于等于(2e6)，那么我们可以记录一下每个数的(phi)是多少，然后记录一下每个(phi)出现的次数后就可以转换成上述题意了(这个思想昨天的(CCPC)网络赛1010也用到了，不过这题由于后面不会处理就没补了23333)，假设(c_i)表示(phi=i)的个数，那么求解的式子就变成了
[F(n)=sumlimits_{k=1}^{n}phi(k)sumlimits_{i=1}^{n}sumlimits_{j=1}^{n}c_ic_j[gcd(i,j)=k]
]
令(f(n))为(sumlimits_{i=1}^{n}sumlimits_{j=1}^{n}c_ic_j[gcd(i,j)=n])，(g(n))为(为和的倍数sumlimits_{i=1}^{n}sumlimits_{j=1}^{n}c_ic_j[gcd(i,j) ext{为}n ext{和}n ext{的倍数}])，则
[egin{aligned}
g(n)=&sumlimits_{n|i}sumlimits_{n|j}c_ic_j&\
=&sumlimits_{i=1}^{n}c_isumlimits_{j=1}^{n}c_J&\
=&(sumlimits_{i=1}^{n}c_i)^2&\
=&sumlimits_{n|d}f(d)&
end{aligned}
]
那么
[f(n)=sumlimits_{n|d}mu(frac{d}{n})g(d)
]
最后答案为
[F(n)=sumlimits_{k=1}^{n}phi(k)f(k)
]
然后暴力枚举(k)求解即可。
代码
#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define lson (rt<<1),L,mid
#define rson (rt<<1|1),mid + 1,R
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********
")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("/home/dillonh/CLionProjects/Dillonh/in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)
const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 2000000 + 2;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
bool v[maxn];
int t, n, cnt;
int F[maxn], c[maxn], phi[maxn], p[maxn], mu[maxn];
void init() {
phi[1] = mu[1] = 1;
for(int i = 2; i < maxn; ++i) {
if(!v[i]) {
p[cnt++] = i;
phi[i] = i - 1;
mu[i] = -1;
}
for(int j = 0; j < cnt && i * p[j] < maxn; ++j) {
v[i*p[j]] = 1;
phi[i*p[j]] = phi[i] * (p[j] - 1);
mu[i*p[j]] = -mu[i];
if(i % p[j] == 0) {
mu[i*p[j]] = 0;
phi[i*p[j]] = phi[i] * p[j];
break;
}
}
}
}
int main() {
#ifndef ONLINE_JUDGE
FIN;
time_t s = clock();
#endif
init();
scanf("%d", &t);
while(t--) {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) {
++c[phi[i]];
}
F[1] = c[1];
for(int i = 2; i <= n; ++i) {
F[1] += c[i];
for(int j = 1; j <= n / i; ++j) {
F[i] += c[i*j];
}
}
LL sum = 1LL * F[1] * F[1], ans = 0;
for(int i = 2; i <= n; ++i) {
sum += 1LL* F[i] * F[i] * mu[i];
LL tmp = 0;
for(int j = 1; j <= n / i; ++j) {
tmp += 1LL * F[i*j] * F[i*j] * mu[j];
}
ans += tmp * phi[i];
}
ans += sum;
printf("%lld
", ans);
for(int i = 1; i <= n; ++i) F[i] = c[i] = 0;
}
#ifndef ONLINE_JUDGE
printf("It costs %.3fs
", 1.0 * (clock() - s) / CLOCKS_PER_SEC);
#endif
return 0;
}