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  • Codeforces 997 C

    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
    ", (res + ans) % MOD);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/widsom/p/9306316.html
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