BZOJ5323 [Jxoi2018]游戏【数论/数学】

转载于 2018-05-24 12:28:00 发布 · 62 阅读

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原文链接：http://www.cnblogs.com/Mychael/p/9082269.html

题目链接

BZOJ5323

题解

有一些数是不能被别的数筛掉的
这些数出现最晚的位置就是该排列的\(t(p)\)
所以我们只需找出所有这些数，线性筛一下即可，设有\(m\)个
然后枚举最后的位置
\[ans = \sum\limits_{i = m}^{n} m!(n - m)!{i - 1 \choose m - 1}i\]

复杂度\(O(n)\)

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn = 10000005,P = 1000000007;
int p[maxn],isn[maxn],pi,L,R,len;
int fac[maxn],fv[maxn],inv[maxn];
void pre(){
    fac[0] = fac[1] = inv[0] = inv[1] = fv[0] = fv[1] = 1;
    for (register int i = 2; i <= len; i++){
        fac[i] = 1ll * fac[i - 1] * i % P;
        inv[i] = 1ll * (P - P / i) * inv[P % i] % P;
        fv[i] = 1ll * fv[i - 1] * inv[i] % P;
    }
}
void init(){
    for (register int i = 2; i <= R; i++){
        if (!isn[i]) p[++pi] = i;
        for (register int j = 1; j <= pi && i * p[j] <= R; j++){
            isn[i * p[j]] = p[j];
            if (i % p[j] == 0) break;
        }
        
    }
}
int main(){
    scanf("%d%d",&L,&R); len = R - L + 1;
    pre();
    int cnt = 0;
    if (L == 1) cnt = 1;
    else{
        init();
        for (register int i = L; i <= R; i++){
            if (!isn[i] || i / isn[i] < L)
                cnt++;
        }
    }
    int ans = 0;
    for (register int i = cnt; i <= len; i++){
        ans = (ans + 1ll * fac[i - 1] % P * fv[i - cnt] % P * i % P) % P;
    }
    ans = 1ll * ans * cnt % P * fac[len - cnt] % P;
    printf("%d\n",ans);
    return 0;
}

转载于:https://www.cnblogs.com/Mychael/p/9082269.html