Codeforces 997 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); //head const 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; }