hdu 6053 TrickGCD 容斥 莫比乌斯

比赛没搞出来。不知道借助莫比乌斯函数容斥。菜。
枚举除数时，除数相同的为一块，还可以优化。

#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
const int MAXN = 1e5 + 10;
const ll md = 1e9 + 7;

bool check[MAXN + 10];
int prime[MAXN + 10];
int tot = 0;
int mu[MAXN + 10];
void Moblus()
{
    memset(check, 0, sizeof(check));
    mu[1] = 1;
    tot = 0;
    for (int i = 2; i <= MAXN; i++){
        if (!check[i]){
            prime[tot++] = i;
            mu[i] = -1;
        }
        for (int j = 0; j < tot; j++){
            if (i*prime[j] > MAXN)break;
            check[i*prime[j]] = true;
            if (i%prime[j] == 0){
                mu[i*prime[j]] = 0;
                break;
            }
            else{
                mu[i*prime[j]] = -mu[i];
            }
        }
    }
} //线性筛求Moblus函数
int n, a[MAXN];
ll sum[MAXN * 2];
void init()
{
    memset(sum, 0, sizeof(sum));
}
ll mypow(ll b, int k)
{
    ll ret = 1;
    while (k){
        if (k & 1)
            ret *= b;
        ret %= md;
        b *= b;
        b %= md;
        k >>= 1;
    }
    return ret;
}
int main()
{
    Moblus(); 
    int T;
    scanf("%d", &T);
    for (int kase = 1; kase <= T; kase++){
        init();
        int mx = -1e9, mi = 1e9;
        scanf("%d", &n);
        for (int i = 1; i <= n; i++){
            scanf("%d", &a[i]);
            mx = max(mx, a[i]);
            mi = min(mi, a[i]);
            sum[a[i]]++;
        }
        if (n == 1){
            printf("Case #%d: %lld\n", kase, (ll)a[1] - 1);
            continue;
        }
        for (int i = mi; i < MAXN * 2; i++){
            sum[i] += sum[i - 1];
        }
        ll ans = 0;
        for (int i = 2; i <= 100000; i++){   //枚举除数
            if (i > mi)break;
            if (mu[i] == 0)continue;
            ll fuhao = -mu[i];
            int p = i;
            ll cnt = 1;
            while (p <= mx){    
                int tmp = sum[p + i - 1] - sum[p - 1]; //除数相同的块
                cnt *= mypow(p / i, tmp);
                cnt %= md;
                p += i;
            }
            ans += fuhao*cnt;
            ans = (ans + md) % md;    //wa1
        }
        printf("Case #%d: %lld\n", kase, ans);
    }
    return 0;
}