Codeforces Round #175 (Div. 2) A~D 题解

最新推荐文章于 2023-02-04 15:01:53 发布
最新推荐文章于 2023-02-04 15:01:53 发布
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原文链接：http://www.cnblogs.com/zxyqzy/p/10357142.html
本文提供了四道算法竞赛题目解析，包括构建特定排列、查找大理石、建立排列及排列求和问题，涉及数学组合、搜索和贪心算法等核心概念。

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A.Slightly Decreasing Permutations

Permutation p is an ordered set of integers p1,  p2,  ...,  pn, consisting of n distinct positive integers, each of them doesn't exceed n. We'll denote the i-th element of permutation p as pi. We'll call number n the size or the length of permutation p1,  p2,  ...,  pn.

The decreasing coefficient of permutation p1, p2, ..., pn is the number of such i (1 ≤ i < n), that pi > pi + 1.

You have numbers n and k. Your task is to print the permutation of length n with decreasing coefficient k.

Input

The single line contains two space-separated integers: n, k (1 ≤ n ≤ 105, 0 ≤ k < n) — the permutation length and the decreasing coefficient.

Output

In a single line print n space-separated integers: p1, p2, ..., pn — the permutation of length n with decreasing coefficient k.

If there are several permutations that meet this condition, print any of them. It is guaranteed that the permutation with the sought parameters exists.

Examples
Input
Copy 
5 2
Output
Copy 
1 5 2 4 3
Input
Copy 
3 0
Output
Copy 
1 2 3
Input
Copy 
3 2
Output
Copy 
3 2 1
构造题；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 2000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 100000007;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

int n, k;

int main()
{
	ios::sync_with_stdio(0);
	cin >> n >> k;
	if (k == 0) {
		for (int i = 1; i <= n; i++)cout << i << ' ';
		return 0;
	}
	
	for (int i = 1; i < n - k; i++) {
		cout << i << ' ';
	}
	for (int i = n; i >= n - k; i--)cout << i << ' ';

	return 0;
}

 

B. Find Marble
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Petya and Vasya are playing a game. Petya's got n non-transparent glasses, standing in a row. The glasses' positions are indexed with integers from 1 to n from left to right. Note that the positions are indexed but the glasses are not.

First Petya puts a marble under the glass in position s. Then he performs some (possibly zero) shuffling operations. One shuffling operation means moving the glass from the first position to position p1, the glass from the second position to position p2 and so on. That is, a glass goes from position i to position pi. Consider all glasses are moving simultaneously during one shuffling operation. When the glasses are shuffled, the marble doesn't travel from one glass to another: it moves together with the glass it was initially been put in.

After all shuffling operations Petya shows Vasya that the ball has moved to position t. Vasya's task is to say what minimum number of shuffling operations Petya has performed or determine that Petya has made a mistake and the marble could not have got from position s to position t.

Input

The first line contains three integers: n, s, t (1 ≤ n ≤ 105; 1 ≤ s, t ≤ n) — the number of glasses, the ball's initial and final position. The second line contains n space-separated integers: p1, p2, ..., pn (1 ≤ pi ≤ n) — the shuffling operation parameters. It is guaranteed that all pi's are distinct.

Note that s can equal t.

Output

If the marble can move from position s to position t, then print on a single line a non-negative integer — the minimum number of shuffling operations, needed to get the marble to position t. If it is impossible, print number -1.

Examples
Input
Copy 
4 2 1
2 3 4 1
Output
Copy 
3
Input
Copy 
4 3 3
4 1 3 2
Output
Copy 
0
Input
Copy 
4 3 4
1 2 3 4
Output
Copy 
-1
Input
Copy 
3 1 3
2 1 3
Output
Copy 
-1
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 100000007;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

int n, s, t;
int p[maxn];

int main()
{
//	ios::sync_with_stdio(0);
	n = rd(); s = rd(); t = rd();
	for (int i = 1; i <= n; i++)p[i] = rd();
	if (s == t) {
		puts("0"); return 0;
	}
	bool fg = 1;
	int cnt = 0;
	int pos = s;
	while (1) {
		pos = p[pos]; cnt++;
		if (pos == s) {
			fg = 0; break;
		}
		else if (pos == t) {
			break;
		}
	}
	if (fg == 0)cout << -1 << endl;
	else {
		cout << cnt << endl;
	}
	return 0;
}

 

C. Building Permutation
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Permutation p is an ordered set of integers p1,  p2,  ...,  pn, consisting of n distinct positive integers, each of them doesn't exceed n. We'll denote the i-th element of permutation p as pi. We'll call number n the size or the length of permutation p1,  p2,  ...,  pn.

You have a sequence of integers a1, a2, ..., an. In one move, you are allowed to decrease or increase any number by one. Count the minimum number of moves, needed to build a permutation from this sequence.

Input

The first line contains integer n (1 ≤ n ≤ 3·105) — the size of the sought permutation. The second line contains n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109).

Output

Print a single number — the minimum number of moves.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples
Input
Copy 
2
3 0
Output
Copy 
2
Input
Copy 
3
-1 -1 2
Output
Copy 
6
Note

In the first sample you should decrease the first number by one and then increase the second number by one. The resulting permutation is (2, 1).

In the second sample you need 6 moves to build permutation (1, 3, 2).

贪心地从小到大排序；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 500005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 100000007;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

int n;
int a[maxn];

int main()
{
//	ios::sync_with_stdio(0);
	n = rd();
	for (int i = 1; i <= n; i++)a[i] = rd();
	ll tot = 0;
	sort(a + 1, a + 1 + n);
	for (int i = 1; i <= n; i++) {
		tot += 1ll * abs(i - a[i]);
	}
	cout << (ll)tot << endl;
	return 0;
}

 

D. Permutation Sum
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Permutation p is an ordered set of integers p1,  p2,  ...,  pn, consisting of n distinct positive integers, each of them doesn't exceed n. We'll denote the i-th element of permutation p as pi. We'll call number n the size or the length of permutation p1,  p2,  ...,  pn.

Petya decided to introduce the sum operation on the set of permutations of length n. Let's assume that we are given two permutations of length n: a1, a2, ..., an and b1, b2, ..., bn. Petya calls the sum of permutations a and b such permutation c of length n, where ci = ((ai - 1 + bi - 1) mod n) + 1 (1 ≤ i ≤ n).

Operation means taking the remainder after dividing number x by number y.

Obviously, not for all permutations a and b exists permutation c that is sum of a and b. That's why Petya got sad and asked you to do the following: given n, count the number of such pairs of permutations a and b of length n, that exists permutation c that is sum of a and b. The pair of permutations x, y (x ≠ y) and the pair of permutations y, x are considered distinct pairs.

As the answer can be rather large, print the remainder after dividing it by 1000000007 (109 + 7).

Input

The single line contains integer n (1 ≤ n ≤ 16).

Output

In the single line print a single non-negative integer — the number of such pairs of permutations a and b, that exists permutation c that is sum of a and b, modulo 1000000007 (109 + 7).

Examples
Input
Copy 
3
Output
Copy 
18
Input
Copy 
5
Output
Copy 
1800

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 500005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;

inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}


ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

ll ans[20];
ll fac[20];
ll a[20], b[20];
ll res;
bool use[20], vis[20];
int n;
void dfs(int pos,int up) {
	if (pos == up + 1) {
		res++;
	}
	for (int i = 1; i <= up; i++) {
		if (use[i])continue;
		int C = (a[pos] + i - 2) % up + 1;
		if (vis[C])continue;
		use[i] = 1; vis[C] = 1;
		dfs(pos + 1, up);
		use[i] = 0; vis[C] = 0;
	}
}


int main()
{
//	ios::sync_with_stdio(0);
	n = rd();
	fac[0] = 1ll;
	for (int i = 1; i <= 16; i++)fac[i] = (i*fac[i - 1]) % mod;
/*	for (int j = 1; j <= 16; j++) {
		for (int i = 1; i <= j; i++)a[i] = i;
		ms(vis); ms(use);
		res = 0; dfs(1, j);
		printf("j=%d , ans=%d\n ", j, res);
	}
	*/
	ans[1] = 1; ans[2] = 0; ans[3] = 3;
	ans[4] = 0; ans[5] = 15; ans[6] = 0; ans[7] = 133;
	ans[8] = 0; ans[9] = 2025; ans[10] = 0;
	ans[11] = 37851; ans[12] = 0; ans[13] = 1030367;
	ans[14] = 0; ans[15] = 36362925; ans[16] = 0;
	cout << (ll)ans[n] * fac[n] % mod << endl;
	
	return 0;
}

 



 

转载于:https://www.cnblogs.com/zxyqzy/p/10357142.html

                

        




    

    
  



    



                



	

		

			

				
					
Codeforces Round #834 (Div. 3) A~E题解

				
			

			

				

					weixin_63927680的博客
				

				

					11-19
					
					1791
					
				

			

		

		

			
				
题目：D. Make It Round

题意：

    给定一个n和m，k的取值范围在1~m,求出最大的n*k并满足有最多的末尾0。首先10 = 2*5,所以要有末尾0，首先必须出现2和5，所以我们先分解n，看一下n有多少个5，多少个2，假设n分解之后有4个2，1个5，它本身1个2和1个5相乘等于10，所以剩下3个2，那么此时，k就先补3个5，既5*5*5 = 125，如果此时k还没有超出范围的话，就继续乘10，如果乘10后超出范围了，就在1~9之间相乘，找到最大的那个k。如果k在补5的时候超出

			
		

	



                

            
              



	

		

			

				
					
Codeforces Round 956 (Div. 2) 部分题解A~C

				
			

			

				

					_Djhhh` blog
				

				

					07-08
					
					910
					
				

			

		

		

			
				
给定操作，可以选择一个矩形的一条对角线上的两个端点，让他们加1/2；另一条对角线的端点值加2/1，然后对这些值取模3。考虑网格的边上的值，边上的值的变化会受到其他的影响，所以我们优先去找如何让边上的值相等即可。所有j的aj之和可以被k整除，其中j是k的倍数，k的取值为1~n。问如何分配蛋糕，使三个人获得的蛋糕价值均高于蛋糕价值和的三分之一。给定a和b两个网格，网格中只存在0、1、2三种值。三个人吃蛋糕，每个人对蛋糕不同位置有自己的估价。正扫一遍，反扫一遍，然后比较数组即可。分类讨论：三个人拿蛋糕的顺序。

			
		

	



              


  
              

                


	

		

			

				
					
Codeforces Round #175 (Div. 2)

				
			

			

				

					weixin_34110749的博客
				

				

					04-15
					
					104
					
				

			

		

		

			
				
D题， 纠结， 一开始想用组合数学的方法可是想不出怎么排列组合， 然后又准备暴力结果发现普通暴力在n=15的时候就不知道要运行多少时间。。。
于是去学习了一个 中途相遇的 搜索算法，快了很多， n=15的那组数据也可以在20s内跑完.
其实这个算法就是将 n^n 的复杂度变成了 2*n^(n/2) ,  可以快很多....
本质就是将整个搜索过程分成两半来搜。
 

D. Permut...

			
		

	





	

		

			

				
					
Codeforces Round #175 (Div. 2)---A. Slightly Decreasing Permutations

				
			

			

				

					sxk
				

				

					10-30
					
					951
					
				

			

		

		

			
				
Slightly Decreasing Permutations


time limit per test
2 seconds


memory limit per test
256 megabytes


input
standard input


output
standard output



Permutation p is an ordere

			
		

	



		

			
		



	

		

			

				
					
codeforces 285(Round #175)题解

				
			

			

				

					mlzmlz95的专栏
				

				

					03-25
					
					885
					
				

			

		

		

			
				
A:直接写降序的即可，剩下全部升序
B:水模拟
C:排序+贪心即可
D:先弄一个初始排列a[i]=i，然后枚举排列使其满足要求，可以用stl的next_permutation函数，很方便。然后将a[i]=i这种排列对应的方案数*n!（1~n的排列为n!种，b数组随a对应改变即可）即可，n
E:递推好题！
|p[i]-i|=1,意味着第i位置要是cool位置的话只能取i+1或者i-1，这点

			
		

	





	

		

			

				
					
Codeforces Round #849 (Div. 4)题解A~G2

				
			

			

				

					jiajajia的博客
				

				

					02-04
					
					715
					
				

			

		

		

			
				
蒟蒻初学

			
		

	





	

		

			

				
					
Codeforces Round #838 (Div. 2)题解

				
			

			

				

					Kicamon的博客
				

				

					12-16
					
					735
					
				

			

		

		

			
				
这场的状态不太好。

			
		

	





	

		

			

				
					
Codeforces Round #829 (Div. 2)部分题解A B C1 E

				
			

			

				

					weixin_63927680的博客
				

				

					10-24
					
					1039
					
				

			

		

		

			
				
​Codeforces Round #829 (Div. 2)部分题解A B C1 E
 原题地址：Dashboard - Codeforces Round #829 (Div. 2) - Codeforces     
A. Technical Support
题意：

    有一个只包含“Q”和"A"的字符串，Q代表问题，A代表回答，有问必有答，也就是Q后面一定有n个A（n>=0）

B. Kevin and Permutation
题意：

    给定

​

			
		

	





	

		

			

				
					
Summer Individual - 17 -A

				
			

			

				

					BePosit的博客
				

				

					08-01
					
					229
					
				

			

		

		

			
				
A permutation is a sequence of integers p1, p2, ..., pn, consisting of n distinct positive integers, each of them doesn't exceed n. Let's denote the i-th element of permutation p as pi. We'll call num...

			
		

	





	

		

			

				
					
B. Find Marble

				
			

			

				

					无知的我
				

				

					04-06
					
					1481
					
				

			

		

		

			
				
time limit per test
2 seconds


memory limit per test
256 megabytes


input
standard input


output
standard output



Petya and Vasya are playing a game. Petya's got n non-transparen

			
		

	





	

		

			

				
					
Find Marble

				
			

			

				

					weixin_44824383的博客
				

				

					08-21
					
					211
					
				

			

		

		

			
				
Find Marble
Petya and Vasya are playing a game. Petya’s got n non-transparent glasses, standing in a row. The glasses’ positions are indexed with integers from 1 to n from left to right. Note that the...

			
		

	





	

		

			

				
					
AtCoder Regular Contest 080 [CDEF]

				
			

			

				

					weixin_34117211的博客
				

				

					08-07
					
					204
					
				

			

		

		

			
				
C - 4-adjacent
Time limit : 2sec / Memory limit : 256MB
Problem Statement
We have a sequence of length N, a=(a1,a2,…,aN). Each ai is a positive integer.
Snuke's objective is to permute the element in ...

			
		

	





	

		

			

				
					
【CodeForces 602B】G - 一般水的题2-Approximating a Constant Range

				
			

			

				

					dingguayi7025的博客
				

				

					12-15
					
					159
					
				

			

		

		

			
				
Description

When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equili...

			
		

	





	

		

			

				
					
[2018-5-4]BNUZ你们还差得远呢

				
			

			

				

					星澈的博客
				

				

					05-05
					
					277
					
				

			

		

		

			
				
有一次卡再水题的，还是因为小细节问题，感觉还是题打的少？想想还是自己的细节考虑不到位吧。。。想到一个办法，拿本子把每次的错误一条条记下来，以后再有wa的时候，一条条对照排除下错误，不知道有没有用。然后还要多背单词了，现在读题速度又很慢，还是因为大多单词不认识，加油！！Building PermutationTime Limit:2000MS     Memory Limit:262144KB   ...

			
		

	





	

		

			

				
					
Coder-Strike 2014 - Round 1（A～E）

				
			

			

				

					weixin_30628077的博客
				

				

					04-23
					
					223
					
				

			

		

		

			
				
题目链接

A. Poster
time limit per test:1 secondmemory limit per test:256 megabytesinput:standard inputoutput:standard output


The R1 company has recently bought a high rise building in the centre ...

			
		

	





	

		

			

				
					
Perfect Permutation

				
			

			

				

					无限迭代中......
				

				

					10-04
					
					516
					
				

			

		

		

			
				
Description
A permutation is a sequence of integers p1, p2, …, pn, consisting of n distinct positive integers, each of them doesn’t exceed n. Let’s denote the i-th element of permutation p as pi. We’l...

			
		

	





	

		

			

				
					
双指针法 Codeforces Round #138 (Div. 2) B题 224B

				
			

			

				

					不知所云的博客
				

				

					02-28
					
					356
					
				

			

		

		

			
				
Array
You’ve got an array a, consisting of n integers: a1, a2, …, an. Your task is to find a minimal by inclusion segment [l, r] (1 ≤ l ≤ r ≤ n) such, that among numbers al,  al + 1,  …,  ar there are...

			
		

	





	

		

			

				
					
Codeforces Round 704 (Div. 2)

					
最新发布

				
			

			

				

					02-28
				

			

		

		

			
				
### 关于Codeforces Round 704 Div. 2 的信息

对于Codeforces Round 704 Div. 2的比赛，虽然未直接提及具体题目解析或参赛体验的内容，但是可以根据平台的一贯风格推测该轮比赛同样包含了多种算法挑战。通常这类赛事会涉及数据结构、动态规划、图论等方面的知识。

考虑到提供的参考资料并未覆盖到此特定编号的比赛详情[^1]，建议访问Codeforces官方网站查询官方题解或是浏览社区论坛获取其他选手分享的经验总结。一般而言，在赛后不久就会有详细的解答发布出来供学习交流之用。

为了帮助理解同类型的竞赛内容，这里提供了一个基于过往相似赛事的例子——如何通过居中子数组特性来解决问题的方法：

```cpp
// 假设有一个函数用于处理给定条件下的数组恢复问题
vector<int> restoreArray(vector<vector<int>>& adjacentPairs) {
    unordered_map<int, vector<int>> adj;
    for (auto& p : adjacentPairs){
        adj[p[0]].push_back(p[1]);
        adj[p[1]].push_back(p[0]);
    }
    
    int start = 0;
    for(auto& [num, neighbors] : adj){
        if(neighbors.size() == 1){
            start = num;
            break;
        }
    }

    vector<int> res(adjacentPairs.size() + 1);
    unordered_set<int> seen;

    function<void(int,int)> dfs = [&](int node, int idx){
        seen.insert(node);
        res[idx] = node;
        
        for(auto next : adj[node]){
            if(!seen.count(next)){
                dfs(next, idx + 1);
            }
        }
    };

    dfs(start, 0);

    return res;
}
```

上述代码展示了利用深度优先搜索(DFS)重建原始序列的一种方式，这与某些情况下解决Codeforces比赛中遇到的问题思路相吻合[^4]。

#### 注意事项
由于缺乏针对Codeforces Round 704 Div. 2的具体材料支持，以上解释更多依赖于对同类活动的理解以及编程技巧的应用实例来进行说明。

			
		

	



              




          




        



    





    



      

      

        
      

      





	

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