Discrete Logarithm is a Joke _

思路：

这段代码只能C++ 17的64位版本或者20的64位版本，否则会CE
ai=gai+1a_i = g^{a_{i+1}}ai=gai+1，且样例给了上界的答案，所以倒推
long long会爆，所以要用__int128
__int128的输入输出方式
long long x = 1e18 + 31 cout x会得到1000(省略0），也就是说爆了，
double x = 1e18 + 31 cout x会得到1e+18，printf x会得到1000000000000000000.000000

#include <iostream>using namespace std;typedef __int128 ll;const int N = 1e6 + 10;//const ll mod = 1e18 + 31;   wa!!不要用浮点数表示整数
const ll mod = 1000000000000000031;ll f[N];ll qmi(ll a, ll b, ll q)
{ll res = 1 % q;while (b){if (b & 1) res = res * a % q;a = a * (ll)a % q;b >>= 1;}return res;
}inline __int128 read()
{__int128 x = 0, f = 1;char ch = getchar();while (ch < '0' || ch > '9'){if (ch == '-') f = -1;ch = getchar();}while (ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}return x * f;
}inline void print(__int128 x)
{if (x < 0){putchar('-');x = -x;}if (x > 9)print(x / 10);putchar(x % 10 + '0');
}int main()
{int n = 1e6;f[n] = 300;for (int i = n - 1; i >= 0; i -- )f[i] = qmi(42ll, f[i + 1], mod);int x;cin >> x;print(f[x]);return 0;
}

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