P4139TriangleXor

数学 #计数 #容斥

分为 $4$ 个部分计算

1. 上面的按奇偶性分类
2. 下面的按每一层容斥，即 $siz_i-2\times siz_{i-1}+siz_{i-2}$
3. 左右直接对每一块计算面积

// Author: xiaruize
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;
 
int n;
double res = 0;
 
double calc(int i)
{
    return (double)i / ((double)n + (double)i) * (double)i / 2.0;
}
 
void solve()
{
    cin >> n;
    for (int i = 2; i <= n; i += 2)
    {
        res = res + (double)(n - i + 1) * (calc(i) - calc(i - 1) * 2 + calc(i - 2));
    }
    // debug(res);
    rep(i, 1, (n + 1) / 2)
    {
        res += ((double)n / 4) * (((double)(i * 2 - 1) / ((double)n + (double)(i * 2 - 1)) - (double)(i * 2 - 2) / ((double)n + (double)(i * 2 - 2)))) * 2.0 * 2.0;
    }
    if (n % 2 == 0)
        res += (double)n / 4.0;
    cout << (int)floor(res) << endl;
}
 
#ifndef ONLINE_JUDGE
bool end_of_memory_use;
#endif
 
signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int testcase = 1;
    // cin >> testcase;
    while (testcase--)
        solve();
#ifndef ONLINE_JUDGE
    cerr << "Memory use:" << (&end_of_memory_use - &start_of_memory_use) / 1024.0 / 1024.0 << "MiB" << endl;
    cerr << "Time use:" << (double)clock() / CLOCKS_PER_SEC * 1000.0 << "ms" << endl;
#endif
    return 0;
}