题目链接：https://acm.hdu.edu.cn/showproblem.php?pid=6956
题目大意：n个人传球，球最开始在第1个人手里，接下来每秒拿球的人必须传给另一个人，记合法方案为最后求传给第1个人，第i秒的合法方案数为f(i),现在知道了方案数，求最小时间t使得f(t) = x (mod 998244353)
题目思路：通过递推得出
\[f\left ( i \right )=\left ( n-2 \right )*f\left ( i-1 \right )+\left ( n-1 \right )*f\left ( i-2 \right )
\]
再通过特征方程求解数列的通项公式得出
\[f\left ( i \right )=\frac{ (n-1) *(-1)^{i}+ (n-1)^i}{n}\equiv x
\]
感谢这篇博客提供的解法：https://www.cnblogs.com/ZX-GO/p/15039734.html
AC代码：
#include <unordered_map>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <string>
#include <stack>
#include <deque>
#include <queue>
#include <cmath>
#include <map>
#include <set>
using namespace std;
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 10, M = 1e7 + 30;
const int base = 1e9;
const int P = 131;
const int mod = 998244353;
const double eps = 1e-8;
const double PI = acos(-1.0);
ll hs[N], head[N], nexts[N], id[N], top;
void insert(ll x, ll y, ll mod) //mod传 N
{
ll k = x % mod;
hs[top] = x;
id[top] = y;
nexts[top] = head[k];
head[k] = top++;
}
ll find(ll x, ll mod)
{
ll k = x % mod;
for (int i = head[k]; i != -1; i = nexts[i])
if (hs[i] == x)
return id[i];
return -1;
}
ll gcd(ll a, ll b)
{
return b == 0 ? a : gcd(b, a % b);
}
ll BSGS(ll a, ll b, ll p, ll a1) //质数传1
{
memset(head, -1, sizeof(head));
top = 1;
a %= p, b %= p;
if (a1 == 1 && 1 % p == b % p)
return 0;
if (a == 0)
{
if (b == 0)
return 1;
else
return -1;
}
//unordered_map<ll, ll> hash; //map unordered_map 都试试
ll k = sqrt(p) + 1;
ll ak = 1;
for (ll i = 0; i < k; ++i)
{
ll t = ak * b % p;
insert(t, i, N);
//hash[t] = i;
ak = ak * a % p;
}
for (ll i = 0; i <= k; ++i)
{
/* if (hash.count(a1) && i * k - hash[a1] >= 0)
return i * k - hash[a1]; */
ll j = find(a1, N);
if (j != -1 && i * k - j >= 0)
return i * k - j;
a1 = a1 * ak % p;
}
return -1;
}
ll exBSGS(ll a, ll b, ll p)
{
a %= p, b %= p;
if (b == 1 || p == 1)
return 0;
ll cnt = 0, a1 = 1;
ll d = gcd(a, p);
while (d > 1)
{
if (b % d)
return -1;
p /= d;
b /= d;
a1 = (a1 * a / d) % p;
++cnt;
if (b == a1)
return cnt;
d = gcd(a, p);
}
ll res = BSGS(a, b, p, a1);
if (res == -1)
return -1;
else
return res + cnt;
}
int main()
{
int T;
scanf("%d", &T);
while (T--)
{
ll n, x;
scanf("%lld%lld", &n, &x);
ll ans1 = exBSGS((n - 1), n * x + (n - 1), mod);
if (ans1 % 2 == 0 || ans1 == -1)
ans1 = INF;
ll ans2 = exBSGS((n - 1), n * x + (1 - n), mod);
if (ans2 % 2 == 1 || ans2 == -1)
ans2 = INF;
ll ans = min(ans1, ans2);
if (ans == INF)
printf("-1\n");
else
printf("%lld\n", ans);
}
return 0;
}