51nod1237 最大公约数之和 V3

题意：求

解：

最后一步转化是因为phi * I = Id，故Id * miu = phi

第二步是反演，中间省略了几步...

然后就这样A了......最终式子是个整除分块，后面用杜教筛求一下phi前缀和即可。

#include <cstdio>
#include <map>

typedef long long LL;
const int N = 5000010, T = 5000008;
const LL MO = 1000000007;

int p[N], top, phi[N];
LL Phi[N], inv2;
bool vis[N];
std::map<LL, LL> mp;

inline void getp(int n) {
    phi[1] = 1;
    for(int i = 2; i <= n; i++) {
        if(!vis[i]) {
            p[++top] = i;
            phi[i] = i - 1;
        }
        for(int j = 1; j <= top && i * p[j] <= n; j++) {
            vis[i * p[j]] = 1;
            if(i % p[j] == 0) {
                phi[i * p[j]] = phi[i] * p[j];
                break;
            }
            phi[i * p[j]] = phi[i] * (p[j] - 1);
        }
    }
    for(int i = 1; i <= n; i++) {
        Phi[i] = (Phi[i - 1] + phi[i]) % MO;
    }
    return;
}

LL getPhi(LL x) {
    if(x <= 0) return 0;
    if(x <= T) return Phi[x];
    if(mp.count(x)) return mp[x];
    LL ans = (x + 1) % MO * (x % MO) % MO * inv2 % MO;
    for(LL i = 2, j; i <= x; i = j + 1) {
        j = x / (x / i);
        ans -= (j - i + 1) % MO * getPhi(x / i) % MO;
        ans %= MO;
    }
    return mp[x] = (ans + MO) % MO;
}

int main() {
    LL n;
    getp(T);
    inv2 = (MO + 1) / 2;
    scanf("%lld", &n);
    LL ans = 0;
    for(LL i = 1, j; i <= n; i = j + 1) {
        j = n / (n / i);
        LL temp = (n / i) % MO;
        ans += temp * temp % MO * (getPhi(j) - getPhi(i - 1) + MO) % MO;
        ans %= MO;
    }
    printf("%lld
", (ans + MO) % MO);
    return 0;
}