CF # 369 D2 D、E

D，只要抓住每个点只有一个出度，那么图就能分成几个部分，而且可以发现，一个部分最多一个环。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;
#define LL long long
const int MAXN = 200005;
const int MOD = 1e9 + 7;
int nt[MAXN];

int two[MAXN];
int part[MAXN];
int visit[MAXN];
int acyc[MAXN], acyn[MAXN];

int n, cyc, cyn;

void dfs2(int u){
	visit[u] = 3;
	acyc[cyc] ++;
	if(visit[nt[u]] == 3) return ;
	dfs2(nt[u]);
}

void dfs(int u){

	visit[u] = 2;
	if(visit[nt[u]] == 0){
		dfs(nt[u]);
	}
	else if(visit[nt[u]] == 1){
		cyn = part[nt[u]];
	//	visit[nt[u]] = 1;
	}
	else{
		cyc ++;
		cyn = cyc;
		dfs2(nt[u]);
	}
	visit[u] = 1;
	part[u] = cyn;
	acyn[cyn]++;
}

int main(){

	two[0] = 1;
	for(int i =1 ; i < MAXN; i++)
		two[i] = (two[i - 1] * 2) % MOD;

	scanf("%d", &n);
	memset(visit, 0, sizeof(visit));
	memset(part, -1, sizeof(part));
	for(int i = 1; i <= n; i++)
		scanf("%d", &nt[i]);

	cyc = cyn = 0;

	for(int i = 1; i <= n; i++){
		if(visit[i] == 0){
			cyn = 0;
			dfs(i);
		}
	}

	int ans = 1;
	/*
	for(int i = 1; i <= n; i++)
		printf("%d  ", part[i]);
	puts("");

	for(int i = 0; i <= cyc; i++){
		printf("%d %d \n", acyc[i], acyn[i]);
	}*/

	for(int i = 0; i <= cyc; i++){
		if(i == 0 && acyn[i] > 0){
			ans = ((LL)ans * two[acyn[i] - 1])%MOD;
		}
		if(i > 0){
			ans = (LL)ans * (two[acyn[i]] - (LL)2 * two[acyn[i] - acyc[i]]) % MOD;
		}
	}
	if(ans < 0) ans = ((LL)ans + MOD) % MOD;

	printf("%d\n", ans);

	return 0;
}

E，题解在注释

/*****************

设 2^n = m ,分母为 m ^k ,算的分子为 m * (m - 1) *.... * (m - k + 1)
只要两个互质，最后用1减就是了，至于化简的过程，就是对分子提取因子2.

边界 太多，搞了一晚……不划算

***********/

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>

using namespace std;
#define LL long long

const int MOD =1000003;

int mod[MOD + 5];
bool vis[MOD + 5];

LL tail[100];

int quick(int m, LL k){
	int ret = 1;
	int pow = m;
	while(k){
		if(k&1)
		ret = (LL)ret * pow % MOD;
		k >>= 1;
		pow = (LL)pow * pow % MOD;
	}
	return ret;
}

int cal(LL n, int index){

	int m = quick(2, n);
	int ret = 1;
	int counts = 0;

	memset(vis, false, sizeof(vis));

	for(LL i = 1; i <= tail[index]; i += 2){
		if(vis[(((LL)m - i)%MOD + MOD)%MOD]){
			break;
		}
		mod[counts++] = (((LL)m - i)%MOD + MOD) %MOD;
		vis[mod[counts - 1]] = true;
	}

//	cout << counts << endl;

	LL mc = tail[index] / 2  + 1;
	LL cc = mc / counts;

	mc = mc % counts;

	for(int i = 0; i < counts; i++)
		ret = ((LL)ret * mod[i]) % MOD;

	ret = quick(ret, cc);
	for(int i = 0; i < mc; i++){
		ret = ((LL)ret * mod[i]) %MOD;
	}

//	cout <<"endl" << endl;

	return ret;

}

int main(){

	LL n, k, tmp;
	int counts = 0;
	cin >> n >> k;

	double tt = log(k) / log(2) - n;

	if(n < 64){
		if( tt > 1e-8){
			cout << 1 << " " << 1 << endl;
			return 0;
		}
		else if(tt < 1e-8){

			LL ans = 1;

			for(LL i = 1; i <= n; i++)
				ans = ans * 2;

			if(ans > 0 && ans < k){
				cout << 1<< " "<< 1 << endl;
				return 0;
			}

		}
	}

	tmp = k - 1;

	LL upc = 0;

	while(tmp){

		if(tmp % 2 == 0){

			upc += tmp / 2;

			tail[counts++] = tmp - 1;
			tmp = tmp / 2;
		}
		else {
			upc += tmp / 2;
			tail[counts++] = tmp;
			tmp = (tmp - 1) / 2;
		}

	}

//	cout << upc << endl << endl;

	LL k_1 = k - 1 - upc;

	int down = quick(2, n - 1);
	down = quick(down, k - 1);
	down = (LL)down * quick(2, k_1) % MOD;

	int up = 1;

//	cout << down <<endl;

//	cout << "YES" << counts <<endl;

	for(int i = 0; i < counts; i++){
		up = ((LL)up * cal(n - i, i))%MOD;
	}

	up = ((down - up) % MOD + MOD) %MOD;

	cout << up  <<" "<< down << endl;
}