huntian oy (HDU - 6706)
Meaning of the questions:
For \ (n-, A, B \) , seeking \ (f (n, a, b) = \ sum_ {i = 1} ^ n \ sum_ {j = 1} ^ igcd (i ^ aj ^ a, i ^ B ^ BJ) [GCD (I, J) =. 1] \% (+ 10 ^. 9. 7) \) . Ensure \ (GCD (A, B) =. 1 \) . \ (n-<= 1E9 \) .
answer:
- \(gcd(i^a-j^a,i^b-j^b) = i^{gcd(a,b)}-j^{gcd(a,b)}\)
- \ (\ Sum_ {i = 1} ^ ni [gcd (i, n) = 1] = \ frac {n \ phi (n) + [n = 1]} {2} \)
The above two formulas.
\[f(n,a,b)=\sum_{i=1}^n\sum_{j=1}^igcd(i^a-j^a,i^b-j^b)[gcd(i,j)=1]\]
\[=\sum_{i=1}^n\sum_{j=1}^i(i-j)[gcd(i,j)=1]\]
\[=\sum_{i=1}^n(i*\phi(i)-\sum_{j=1}^ij[gcd(i,j)=1])\]
\ [= \ Sum_ {i = 1} ^ n (i * \ phi (i) - \ frac {i * \ phi (i)} {2}) - \ frac {1} {2} (here, to subtract \ frac {1} {2}, is a special case of considering i = 1) \]
\[=\frac{1}{2}(\sum_{i=1}^{n}(i*\phi(i))-1)\]
For \ (\ sum_ {i = 1} ^ {i} the n-* \ Phi (i) \) , can be used to teach Du Du teach sieve sieve seek :( study recommended this blog )
For \ (F (n-) = I * \ Phi (I) \) , configured \ (G (n-) = n-\) . The convolution disposed Deeley Cray \ (F * G = H \) , then
\[h(n)=\sum_{d|n}f(d)*g(\frac{n}{d})\]
\[=\sum_{d|n}(d*\phi(d))*\frac{n}{d}\]
\[=\sum_{d|n}n*\phi(d)\]
\ [= N \ sum_ {d | n} \ phi (d) = n ^ 2 \]
Therefore, in accordance with teachings Du sieve set \ (S (n-) = \ sum_. 1 = {I}} ^ {n-I * \ Phi (I) \) , then
\[S(n)=\sum_{i=1}^{n}i^2-\sum_{d=2}^{n}d*S(n)\]
On the block number of the recursive computation + \ (S (n) \) like.
Code:
#include <bits/stdc++.h>
#define fopi freopen("in.txt", "r", stdin)
#define fopo freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
const int MOD = 1e9+7;
const LL inv2 = 5e8+4;
const LL inv6 = 166666668;
typedef long long LL;
const int maxn = 1e6 + 10;
int check[maxn], phi[maxn], prime[maxn];
LL sum[maxn];
unordered_map<int, LL> d;
void init(int N) {
memset(check, false, sizeof(check));
phi[1] = 1;
int tot = 0;
for (int i = 2; i <= N; i++) {
if (!check[i]) {
prime[tot++] = i;
phi[i] = i-1;
}
for (int j = 0; j < tot; j++) {
if (i * prime[j] > N) break;
check[i * prime[j]] = true;
if (i % prime[j] == 0) {
phi[i * prime[j]] = phi[i] * prime[j];
break;
}
else {
phi[i * prime[j]] = phi[i] * (prime[j]-1);
}
}
}
for (int i = 1; i <= N; i++)
sum[i] = (sum[i-1] + 1ll*i*phi[i] % MOD) % MOD;
}
LL solve(int x) {
return 1ll * x * (x+1) % MOD * (2*x+1) % MOD * inv6 % MOD;
}
int dfs(int x) {
if (x <= 1000000) return sum[x];
if (d[x]) return d[x];
LL ans = solve(x);
for (int l = 2, r; l <= x; l = r+1) {
r = x/(x/l);
ans = (ans - (1LL*(r-l+1)*(l+r)/2) % MOD * dfs(x/l) % MOD + MOD) % MOD;
}
return d[x] = ans;
}
int T;
int n, a, b;
int main() {
init(1000000);
scanf("%d", &T);
for (int ca = 1; ca <= T; ca++) {
scanf("%d%d%d", &n, &a, &b);
printf("%lld\n", inv2 * (dfs(n) - 1 + MOD) % MOD);
}
}