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  • poj 2369 Permutations 置换

    题目链接

    给一个数列, 求这个数列置换成1, 2, 3....n需要多少次。

    就是里面所有小的置换的长度的lcm。

    #include <iostream>
    #include <vector>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <cmath>
    #include <map>
    #include <set>
    #include <string>
    #include <queue>
    #include <stack>
    #include <bitset>
    using namespace std;
    #define pb(x) push_back(x)
    #define ll long long
    #define mk(x, y) make_pair(x, y)
    #define lson l, m, rt<<1
    #define mem(a) memset(a, 0, sizeof(a))
    #define rson m+1, r, rt<<1|1
    #define mem1(a) memset(a, -1, sizeof(a))
    #define mem2(a) memset(a, 0x3f, sizeof(a))
    #define rep(i, n, a) for(int i = a; i<n; i++)
    #define fi first
    #define se second
    typedef pair<int, int> pll;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int mod = 1e9+7;
    const int inf = 1061109567;
    const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
    int a[1005], vis[1005];
    int gcd(int x, int y) {
        return y?gcd(y, x%y):x;
    }
    int lcm(int x, int y) {
        return x/gcd(x, y)*y;
    }
    int main()
    {
        int n;
        while(cin>>n) {
            for(int i = 0; i<n; i++){
                scanf("%d", &a[i]);
                a[i]--;
            }
            int ans = 1;
            for(int i = 0; i<n; i++) {
                if(!vis[i]) {
                    int tmp = i, len = 0;
                    while(!vis[tmp]) {
                        vis[tmp] = 1;
                        len++;
                        tmp = a[tmp];
                    }
                    ans = lcm(ans, len);
                }
            }
            cout<<ans<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/yohaha/p/5257661.html
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