求1-n 中与 m 互质的素因子 （容斥原理）

ll prime[100];
ll cnt;
void getprime(){
    cnt = 0; ll num = m;
    for(ll i = 2; i*i <= m; i++){ // sqrt(m) 的复杂度求出m的素因子
        if (num%i == 0) {
            prime[cnt++] = i;
            while(num%i == 0){
                num /= i;
            }
        }
        if (num == 1) break;
    }
    if (num > 1) prime[cnt++] = num; 
}

void solve() {
    ll ans = 0;
    // cnt 下标从0开始
    for(ll i = 1; i < (1<<cnt); i++){
        ll f = 0; ll tem = 1;
        for(ll j = 0; j < cnt; j++){
            if (i&(1<<j)) {
                f++;
                tem *= prime[j];
            }
        }
        ll time = n/tem;
        // 奇加偶减
        if (f&1) ans = (ans+cal1(tem, time)+cal2(tem, time))%mod;
        else ans = (ans-cal1(tem, time)-cal2(tem, time))%mod;
        ans = (ans+mod)%mod;
    }
    ll sum = (cal1(1, n)+cal2(1, n))%mod; 
    ans = (sum-ans)%mod;
    printf("%lld
", (ans+mod)%mod);
}