[BZOJ 5323][Jxoi2018]游戏

传送门

\(\color{green}{solution}\)

它每次感染的人是它的倍数,那么我们只需要找出那些除了自己以外在\(l\), \(r\)内没有别的数是
它的约数的数,在这里称其为关键数.
(比如区间是[3,7],那么这些数分别是\(3\),\(4\),\(5\),\(7\).因为\(6\)是\(3\)的倍数,所以不是).
我们记这些数的数量为\(sum\).那么设\(f_i\)为在第\(i\)时刻将说有人感染完的方案数.
那么换句话来说,就是最后一个关键数在第\(i\)时刻被感染.
那么\(f_i\) = \(sum\) \(\times (_{n-i}^{n-sum}) \times(i - 1)! \times(n-i)!\).所以最后\(ans = \sum_{i=sum}^{n} i \times f_i\)

BZOJ的垃圾评测机+卡常真的让人难受

#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7, maxn = 1e7 + 10;
int fac[maxn], ifac[maxn], l, r;
int C(int n, int m) {
    if( n < m) return 0;
    return 1LL * fac[n] * ifac[m] % mod * ifac[n-m] % mod;
}
int _pow(int x, int n) {
    int ret = 1;
    for ( ; n; n >>= 1, x = 1LL * x * x % mod)
        if( n & 1) ret = 1LL * ret * x % mod;
    return ret;
}
bool book[maxn];
int pri[maxn], ipri[maxn], cnt;
int sum, ans, n;
int main() {
    scanf("%d%d", &l, &r);
    fac[0] = 1;
    for ( register int i = 1; i <= r; ++ i) {
        fac[i] = 1LL * fac[i-1] * i % mod;
    }
    ifac[r] = _pow(fac[r], mod-2);
    for ( register int i = r-1; ~i; -- i) {
        ifac[i] = 1LL * ifac[i+1] * (i + 1) % mod;
    }
    if( l == 1) {
        ans = 1LL * fac[r] * (r + 1) % mod * ifac[2] % mod;
        printf("%d\n", ans); return 0;
    }
//  for ( register int i = l; i <= r; ++ i) if( !book[i]) {
//      ++ sum;
//      for ( register int k = i; k <= r; k += i) book[k] = 1;
//  }
    for ( register int i = 2; i <= r; ++ i) {
        if( !book[i]) pri[++ cnt] = i, ipri[i] = 1;
        for ( register int k = 1; k <= cnt; ++ k) {
            if( 1LL * pri[k] * i > r) break;
            ipri[i * pri[k]] = i; book[i*pri[k]] = 1;
            if( i % pri[k] == 0) break;
        }
    }
    for ( register int i = l; i <= r; ++ i) {
        if( ipri[i] < l) ++ sum;
    }
    n = r - l + 1;
    for ( register int i = sum; i <= n; ++ i) {
        ans = (ans + 1LL * i * sum % mod * C(n - sum, n - i) % mod * fac[n - i] % mod * fac[i - 1] % mod) % mod;
    }
    printf("%d\n", ans);
    return 0;
}

吐槽:
九条可怜 + 组合数学 = 瞬间爆炸.还有为什么可以二次感染,度错题写了半天QAQ;

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