标签：mod std 题目 const lld ons 方案 i++ 题解
题目大意：给你$n$，一种合法的排列为，排列中没有$s[i\%n+1]-s[i]==1$，求合法方案数
题解：容斥，令$f_{i,j}$表示有$i$个元素，至少包含$j$个$s[i\%n+1]-s[i]==1$的方案数，发现$f_{n,1}=\binom n 1(n-2)!$个
推广$f_{n,k}=\binom n k(n-k-1)!$（令$(-1)!==1$）
$\therefore ans = (-1)^n + \sum_{k = 0}^{n - 1} (-1)^k \binom{n}{k} (n - k - 1)!$
卡点：无
C++ Code：
#include <cstdio>
#define maxn 100010
const long long mod = 998244353;
int Tim, n;
long long FAC[maxn + 1], inv[maxn], *fac = &FAC[1], ans;
long long C(long long a, long long b) {
if (a < b) return 0;
return fac[a] * inv[b] % mod * inv[a - b] % mod;
}
int main() {
scanf("%d", &Tim);
fac[-1] = fac[0] = fac[1] = inv[0] = inv[1] = 1;
for (int i = 2; i <= 100000; i++) {
fac[i] = fac[i - 1] * i % mod;
inv[i] = inv[mod % i] * (mod - mod / i) % mod;
}
for (int i = 2; i <= 100000; i++) inv[i] = inv[i] * inv[i - 1] % mod;
while (Tim --> 0) {
scanf("%d", &n);
ans = 0;
for (int i = 0; i <= n; i++) {
ans = (ans + ((i & 1) ? -1ll : 1ll) * C(n, i) * fac[n - i - 1] % mod) % mod;
}
if (ans < 0) ans += mod;
printf("%lld\n", ans);
}
return 0;
}
标签：mod std 题目 const lld ons 方案 i++ 题解
原文地址：https://www.cnblogs.com/Memory-of-winter/p/9664783.html
