题意:
已知(f(n,a,b)=sum_{i=1}^nsum_{j=1}^igcd(i^a-j^a,i^b-j^b)[gcd(i,j)=1]mod 1e9+7)，(nleq1e9)，且保证(ab)互质，求(f(n,a,b))
思路：
由不知道什么得：当(ab)互质，则(gcd(i^a-j^a,i^b-j^b)=i-j)。
那么可转化为：
[egin{aligned}
&sum_{i=1}^nsum_{j=1}^ii-j[gcd(i,j)=1]mod 1e9+7\
=&sum_{i=1}^nivarphi(i)-frac{i*varphi(i)}{2}\
=&frac{1}{2}(sum_{i=1}^nivarphi(i) - 1)
end{aligned}
]
那么问题转化为求：(sum_{i=1}^nivarphi(i))，因为前缀和有点大，所以估计要用到杜教筛，所以先卷积一下：
[egin{aligned}
(f*g)(n)=&sum_{d|n}f(d)g(frac{n}{d})\
=&sum_{d|n}ivarphi(i)g(frac{n}{d})
end{aligned}
]
为了让卷积完的(h)和(g)前缀更好求，那么这里不如令(g=id)，那么((f*g)(n)=n*sum_{d|n}varphi(d))，又因为(varphi*I=id)，故((f*g)(n)=n^2)。
所以可得终式：
[S(n)=sum_{i=1}^ni^2-sum_{i=2}iS(lfloorfrac{n}{i}
floor)
]
因为超过(1e6)只有(10)组，所以我这里直接手动哈希比(map)快一点。
代码：
#include<map>
#include<set>
#include<cmath>
#include<cstdio>
#include<stack>
#include<ctime>
#include<vector>
#include<queue>
#include<cstring>
#include<string>
#include<sstream>
#include<iostream>
#include<algorithm>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 4e6 + 5;
const ll MOD = 1e9 + 7;
const ull seed = 131;
const int INF = 0x3f3f3f3f;
int mark[maxn], prime[maxn], tot;
ll phi[maxn];
void init(int n){
tot = 0;
int i, j;
phi[1] = 1;
for(i = 2; i <= n; i++){
if(!mark[i]){
prime[++tot] = i;
phi[i] = i - 1;
}
for(j = 1; j <= tot; j++){
if(i * prime[j] > n) break;
mark[i * prime[j]] = 1;
if(i % prime[j] == 0){
phi[i * prime[j]] = phi[i] * prime[j];
break;
}
else phi[i * prime[j]] = phi[i] * (prime[j] - 1);
}
}
phi[0] = 0;
for(int i = 1; i <= n; i++){
phi[i] = (1LL * phi[i] * i + phi[i - 1]) % MOD;
}
}
ll inv6 = 166666668, inv2 = 500000004;
ll mphi[100000], UP;
int vis[100000];
int N = 4e6;
ll getans(int x){
return x <= N? phi[x] : mphi[UP / x];
}
ll ksc(ll a, ll b, ll p){
ll t = a * b - (ll)((long double)a * b / p + 0.5) * p;
return (t < 0)? t + p : t;
}
void solve(int n){
int t = UP / n;
if(n <= N) return;
if(vis[t]) return;
vis[t] = 1;
mphi[t] = ksc(ksc(n, n + 1, MOD), 2 * n + 1, MOD) * inv6 % MOD;
for(int l = 2, r; l <= n; l = r + 1){
r = n / (n / l);
solve(n / l);
mphi[t] -= 1LL * (l + r) * (r - l + 1) % MOD * inv2 % MOD * getans(n / l) % MOD;
mphi[t] %= MOD;
}
}
int main(){
int T;
init(N);
// init(N + 1);
// printf("%lld
", (phi[N + 1] - 1) * inv2 % MOD);
scanf("%d", &T);
while(T--){
ll ans;
int n, a, b;
scanf("%d%d%d", &n, &a, &b);
if(n <= N) ans = phi[n];
else{
UP = n;
memset(vis, 0, sizeof(vis));
solve(n), ans = mphi[1];
}
ans = (ans - 1) * inv2;
ans = (ans % MOD + MOD) % MOD;
printf("%lld
", ans);
}
return 0;
}