Codeforces 997 C - Sky Full of Stars

C - Sky Full of Stars

思路：

容斥原理

题解：http://codeforces.com/blog/entry/60357

注意当i > 1 且 j > 1，是同一种颜色

代码：

#include<iostream>
#include<cstdio>
#include<queue>
#include<deque>
#include<set>
#include<cstring>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//headconst int MOD = 998244353;LL q_pow(LL n, LL k) {LL ans = 1;while(k) {if(k&1) ans = (ans * n) % MOD;n = (n*n) % MOD;k >>= 1;}return ans;
}
int main() {int n;scanf("%d", &n);LL res = 0, t = 1, sign = -1;for (int i = 1; i <= n; i++) {t = (t * (n-i+1)) % MOD;t = (t * q_pow(i, MOD - 2)) % MOD;sign = -sign;LL tt = q_pow(3, 1LL*n*(n-i)+i);res = (res + tt*t*sign) % MOD;}res = (res * 2) % MOD;res = (res + MOD) % MOD;LL ans = 0;t = 1, sign = -1;for (int i = 0; i < n; i++) {LL tt = q_pow( 1 - q_pow(3, i), n);tt = (tt - q_pow( - q_pow(3, i), n)) % MOD;ans = (ans + tt*t*sign) % MOD;t = (t * (n-i)) % MOD;t = (t * q_pow(i+1, MOD-2)) % MOD;sign = -sign;}ans = (ans * 3) % MOD;ans = (ans + MOD) % MOD;printf("%lld\n", (res + ans) % MOD);return 0;
}

转载于:https://www.cnblogs.com/widsom/p/9306316.html

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