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  • [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 = 200 << 20;
    static char buf[maxn], *pbuf = buf, *End;
    void init() {
        int c = fread(buf, 1, maxn, stdin);
        End = buf + c;
    }
    int &readint() {
        static int ans;
        ans = 0;
        while (pbuf != End && !isdigit(*pbuf)) pbuf ++;
        while (pbuf != End && isdigit(*pbuf)) {
            ans = ans * 10 + *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=0;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?1:-1;
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?1:-1;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 + 7;
const double EPS = 1e-8;

/* -------------------------------------------------------------------------------- */

const int maxn = 3e6 + 7;
const unsigned int md = 3221225473;

vector<int> prime;
vector<vector<pii> > R;
bool vis[maxn], flag[maxn];
int power[maxn], a[maxn];

void init() {
    for (ll i = 2; 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 = 0; x > 1 && i < prime.size(); i ++) {
        int c = 0;
        while (x % prime[i] == 0) {
            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 == 0) return 1;
    ull buf = powermod(a, b >> 1, md);
    buf = buf * buf % md;
    if (b & 1) buf = buf * a % md;
    return buf;
}

namespace IO {
    const static int maxn = 200 << 20;
    static char buf[maxn], *pbuf = buf, *End;
    void init() {
        int c = fread(buf, 1, maxn, stdin);
        End = buf + c;
    }
    int &readint() {
        static int ans;
        static char ch;
        ans = 0;
        while (pbuf != End && !isdigit(*pbuf)) pbuf ++;
        while (pbuf != End && isdigit(*pbuf)) {
            ans = ans * 10 + *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 = 1; i <= n; i ++) {
            a[i] = IO::readint();
        }
        fillchar(vis, 0);
        R.clear();
        for (int i = 1; i <= n; i ++) {
            if (vis[i] || a[i] == i) continue;
            int cnt = 0;
            for (int j = i; !vis[j]; j = a[j]) {
                vis[j] = true;
                cnt ++;
            }
            add(cnt);
        }
        fillchar(power, 0);
        int maxpower = 0;
        for (int i = 0; i < R.size(); i ++) {
            for (int j = 0; j < R[i].size(); j ++) {
                umax(power[R[i][j].X], R[i][j].Y);
                umax(maxpower, R[i][j].X);
            }
        }
        unsigned int ans = 1;
        for (int i = 0; i <= maxpower; i ++) {
            ans = ((ull)ans * powermod(prime[i], power[i], md)) % md;
        }
        printf("%u
", ans);
    }
    return 0;
}
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  • 原文地址：https://www.cnblogs.com/jklongint/p/4734253.html
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