huntian oy (HDU - 6706)
题意：
给(n,a,b)，求(f(n,a,b)=sum_{i=1}^nsum_{j=1}^igcd(i^a-j^a,i^b-j^b)[gcd(i,j)=1]\%(10^9+7))。保证(gcd(a,b)=1)。(n<=1e9)。
题解：
- (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{nphi(n)+[n=1]}{2})
根据上面两个公式。
[f(n,a,b)=sum_{i=1}^nsum_{j=1}^igcd(i^a-j^a,i^b-j^b)[gcd(i,j)=1]
]
[=sum_{i=1}^nsum_{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}（这里要减去frac{1}{2}，是特殊考虑了i=1的情况）
]
[=frac{1}{2}(sum_{i=1}^{n}(i*phi(i))-1)
]
对于(sum_{i=1}^{n}i*phi(i))，可以用杜教筛求：(杜教筛学习推荐这篇博客)
对于(f(n)=i*phi(i))，构造(g(n)=n)。根据迪利克雷卷积设(h=f*g)，则
[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)
]
[=nsum_{d|n}phi(d)=n^2
]
所以根据杜教筛，设(S(n)=sum_{i=1}^{n}i*phi(i))，则
[S(n)=sum_{i=1}^{n}i^2-sum_{d=2}^{n}d*S(n)
]
数论分块+递归计算(S(n))就好。
代码：
#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
", inv2 * (dfs(n) - 1 + MOD) % MOD);
}
}