HDU 6038 Function （2017多校1 -- 找循环节计数）-CFANZ编程社区

题意：

给你a数组和b数组，问你f 函数的种类。使得a[i] = b[f[a[i]]]？

思路：

因为a 数组和b 数组都是0~n-1的排列。

所以一定有一个循环节：

比如 2 1 0 3的话，就有两个循环节。

2 1 0 和 3

那么要想使得是一个循环的a 赋值，就也得找b 的循环节。

假如我们得到了b的循环节。

那么对于每一个a 来说，只要b 是a 的因子，那么b就可以给a 赋值。方案数是b 循环节的长度。

那么把所有的乘起来即可。

无力吐槽：

因为初始化数组是循环跑的，忘记了n 和m 不一定相等， wa了一天。。。。。。。。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;

const int maxn = 100000 + 10;

int a[maxn];
int b[maxn];

int sum[maxn];

int vis[maxn];

vector<int>va, vb;

void dfs(int cur,int c[],int n,int ans,int flag){
    vis[cur] = 1;
    if (vis[c[cur] ]) {
        if (flag) {
            sum[ans]++;
        }
        else va.push_back(ans);
    }
    else {
        dfs(c[cur], c, n, ans+1, flag);

    }
}

const int mod = 1e9 + 7;
void add(long long& ans, long long x){
    ans += x;
    if (ans >= mod) ans -= mod;
}
int main(){
    int n,m;
    int ks = 0;
    while(~scanf("%d %d",&n, &m)){
        va.clear();
        for (int i = 0; i < n; ++i){
            scanf("%d", a+i);
        }
        memset(sum,0,sizeof sum); /// 醉了= =
        for (int i = 0; i < m; ++i){
            scanf("%d",&b[i]);
        }


        memset(vis,0,sizeof vis);
        for (int i = 0; i < n; ++i){
            if (!vis[i]){
                dfs(i,a, n, 1, 0);
            }

        }

        memset(vis,0,sizeof vis);
        for (int i = 0; i < m; ++i){
            if (!vis[i]){
                dfs(i,b, m, 1, 1);
            }
        }

        long long ans = 1LL;
        for (int i = 0; i < va.size(); ++i){
            int x = va[i];
            int mm = sqrt(x + 0.5);
            long long tmp = 0LL;
            for (int j = 1; j <= mm; ++j){
                if (x % j == 0){
                    if (j * j != x){
                        if (sum[j])add(tmp, (long long)j * (long long)sum[j]);
                        if (sum[x/j])add(tmp, (long long)(x/j) * (long long)sum[x/j] );
                    }
                    else {
                        if (sum[j])add(tmp, (long long)j * (long long)sum[j]);
                    }
                }
            }
            ans = (ans * tmp) % mod;
        }
        printf("Case #%d: %I64d\n", ++ks,ans);
    }
    return 0;
}
/**
3 2
1 0 2
0 1
3 4
2 0 1
0 2 3 1


Case #1: 4
Case #2: 4

**/

Function Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1351 Accepted Submission(s): 623 Problem Description a from 0 to n−1 and a permutation b from 0 to m−1. Define that the domain of function f is the set of integers from 0 to n−1, and the range of it is the set of integers from 0 to m−1. Please calculate the quantity of different functions f satisfying that f(i)=bf(ai) for each i from 0 to n−1. Two functions are different if and only if there exists at least one integer from 0 to n−1 mapped into different integers in these two functions. The answer may be too large, so please output it in modulo 109+7. Input The input contains multiple test cases. For each case: The first line contains two numbers n, m. (1≤n≤100000,1≤m≤100000) The second line contains n numbers, ranged from 0 to n−1, the i-th number of which represents ai−1. The third line contains m numbers, ranged from 0 to m−1, the i-th number of which represents bi−1. It is guaranteed that ∑n≤106, ∑m≤106. Output Case #x: y" in one line (without quotes), where x indicates the case number starting from 1 and y Sample Input `3 21 0 20 13 42 0 10 2 3 1` Sample Output `Case #1: 4Case #2: 4` Source 2017 Multi-University Training Contest - Team 1 Recommend liuyiding \| We have carefully selected several similar problems for you: 6055 6054 6053 6052 6051 Statistic \| Submit \| Discuss \| Note

Function Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1351 Accepted Submission(s): 623 Problem Description a from 0 to n−1 and a permutation b from 0 to m−1. Define that the domain of function f is the set of integers from 0 to n−1, and the range of it is the set of integers from 0 to m−1. Please calculate the quantity of different functions f satisfying that f(i)=bf(ai) for each i from 0 to n−1. Two functions are different if and only if there exists at least one integer from 0 to n−1 mapped into different integers in these two functions. The answer may be too large, so please output it in modulo 109+7. Input The input contains multiple test cases. For each case: The first line contains two numbers n, m. (1≤n≤100000,1≤m≤100000) The second line contains n numbers, ranged from 0 to n−1, the i-th number of which represents ai−1. The third line contains m numbers, ranged from 0 to m−1, the i-th number of which represents bi−1. It is guaranteed that ∑n≤106, ∑m≤106. Output For each test case, output " Case #x: y" in one line (without quotes), where x indicates the case number starting from 1 and y Sample Input 3 21 0 20 13 42 0 10 2 3 1 Sample Output Case #1: 4Case #2: 4 Source 2017 Multi-University Training Contest - Team 1 Recommend liuyiding \| We have carefully selected several similar problems for you: 6055 6054 6053 6052 6051 Statistic \| Submit \| Discuss \| Note