[hdu5392 Infoplane in Tina Town]置换的最小循环长度，最小公倍数取模，输入挂

题意：给一个置换，求最小循环长度对p取模的结果

思路：一个置换可以写成若干循环的乘积，最小循环长度为每个循环长度的最小公倍数。求最小公倍数对p取模的结果可以对每个数因式分解，将最小公倍数表示成质数幂的乘积形式，然后用快速幂取模，而不能一边求LCM一边取模。

由于这题数据量太大，需要用到输入挂，原理是把文件里面的东西用fread一次性读到内存。

输入挂模板：

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namespace IO { const static int maxn = << ; static char buf[maxn], *pbuf = buf, *End; void init() { int c = fread(buf, , maxn, stdin); End = buf + c; } int &readint() { static int ans; ans = ; while (pbuf != End && !isdigit(*pbuf)) pbuf ++; while (pbuf != End && isdigit(*pbuf)) { ans = ans * + *pbuf - '0'; pbuf ++; } return ans; } }

源程序：

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#pragma comment(linker, "/STACK:10240000") #include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define copy(a, b) memcpy(a, b, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; typedef unsigned long long ull; //#ifndef ONLINE_JUDGE void RI(vector<int>&a,int n){a.resize(n);for(int i=;i<n;i++)scanf("%d",&a[i]);} void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?:-; while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?:-;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} //#endif template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} const double PI = acos(-1.0); const int INF = 1e9 + ; const double EPS = 1e-8; /* -------------------------------------------------------------------------------- */ const int maxn = 3e6 + ; const unsigned int md = ; vector<int> prime; vector<vector<pii> > R; bool vis[maxn], flag[maxn]; int power[maxn], a[maxn]; void init() { for (ll i = ; i < maxn; i ++) { if (flag[i]) continue; prime.pb(i); for (ll j = i * i; j < maxn; j += i) { flag[j] = true; } } } void add(int x) { vector<pii> buf; for (int i = ; x > && i < prime.size(); i ++) { int c = ; while (x % prime[i] == ) { c ++; x /= prime[i]; } if (c) buf.pb(mp(i, c)); } R.pb(buf); } unsigned int powermod(int a, int b, unsigned int md) { if (b == ) return ; ull buf = powermod(a, b >> , md); buf = buf * buf % md; if (b & ) buf = buf * a % md; return buf; } namespace IO { const static int maxn = << ; static char buf[maxn], *pbuf = buf, *End; void init() { int c = fread(buf, , maxn, stdin); End = buf + c; } int &readint() { static int ans; static char ch; ans = ; while (pbuf != End && !isdigit(*pbuf)) pbuf ++; while (pbuf != End && isdigit(*pbuf)) { ans = ans * + *pbuf - '0'; pbuf ++; } return ans; } } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE int T, n; IO::init(); T = IO::readint(); init(); while (T --) { n = IO::readint(); for (int i = ; i <= n; i ++) { a[i] = IO::readint(); } fillchar(vis, ); R.clear(); for (int i = ; i <= n; i ++) { if (vis[i] || a[i] == i) continue; int cnt = ; for (int j = i; !vis[j]; j = a[j]) { vis[j] = true; cnt ++; } add(cnt); } fillchar(power, ); int maxpower = ; for (int i = ; i < R.size(); i ++) { for (int j = ; j < R[i].size(); j ++) { umax(power[R[i][j].X], R[i][j].Y); umax(maxpower, R[i][j].X); } } unsigned int ans = ; for (int i = ; i <= maxpower; i ++) { ans = ((ull)ans * powermod(prime[i], power[i], md)) % md; } printf("%u\n", ans); } return ; }