# Codeforces Round #529(Div.3)个人题解

Codeforces Round #529(Div.3)个人题解

前言： 闲来无事补了前天的cf，想着最近刷题有点点怠惰，就直接一场cf一场cf的刷算了，以后的题解也都会以每场的形式写出来

---

A. Repeating Cipher

传送门

题意：第一个字母写一次，第二个字母写两次，依次递推，求原字符串是什么

题解：1、2、3、4，非常明显的d=1的等差数列，所以预处理一个等差数列直接取等差数列的每一项即可

代码：

#include<bits/stdc++.h>
using namespace std;
int num[100000];
void init(){
	int ans=0;
	for(int i=1;i<=1000;i++){
		ans+=i;
		num[i]=ans;
	}
	return;
}
char str[10000];
int main(){
	int n;
	init();
	scanf("%d %s",&n,str+1);
	int tmp=1;
	while(num[tmp]!=n){
		tmp++;
	}
	for(int i=1;i<=tmp;i++){
		cout<<str[num[i]];
	}
	cout<<endl;
}

B. Array Stabilization

传送门

题意：给你一串数字，要你删除一个数最小化这串数字中最大值-最小值的差

题解：用multiset存一下，然后讨论删去最大的数更好还是删去最小的数更好

代码：

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define PI acos(-1)
#define eps 1e-8
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> PII;
const int maxn = 3e5 + 5;
const LL INF = 1e18 + 7;
const ull mod = 9223372034707292160;
LL gcd(LL a, LL b) {return b ? gcd(b, a % b) : a;}
LL lcm(LL a, LL b) {return a / gcd(a, b) * b;}
LL powmod(LL a, LL b, LL MOD) {LL ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
double dpow(double a, LL b) {double ans = 1.0; while (b) {if (b % 2)ans = ans * a; a = a * a; b /= 2;} return ans;}
multiset<int> s;
int main() {
#ifndef ONLINE_JUDGE
	FIN
#endif
	int n;
	cin >> n;
	int x;
	for (int i = 0; i < n; i++) {
		scanf("%d", &x);
		s.insert(x);
	}
	multiset<int>::iterator it;
	it = s.begin();
	int minn = *it;
	it = s.end();
	it--;
	int maxx = *it;
	if (s.count(minn) != 1 && s.count(maxx) != 1) {
		cout << maxx - minn << endl;
	} else if (s.count(minn) == 1 && s.count(maxx) == 1) {
		it = s.begin();
		it++;
		int tmp1 = *it;
		it = s.end();
		it--;
		it--;
		int tmp2 = *it;
		int ans1 = maxx - tmp1;
		int ans2 = tmp2 - minn;
		cout << min(ans1, ans2) << endl;
	} else {
		if (s.count(minn) == 1) {
			it = s.begin();
			it++;
			cout << maxx - *it << endl;
		} else {
			it = s.end();
			it--;
			it--;
			cout << *it - minn << endl;
		}
	}
}

C. Powers Of Two

传送门

题意：给你一个数n，要求你用k个2的幂次数去拼出这个数，如果不能输出-1

题解：先将n转换为相对应的二进制数 ，如果n的二进制数中的1的个数大于k，显然是没有解的，如果n的二进制数中的1的个数小于1，那么就把每一个大于2的数分解（x->x/2+x/2），凑出k个1即可，最后输出 一下就行

代码

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define PI acos(-1)
#define eps 1e-8
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> PII;
const int maxn = 3e5 + 5;
const LL INF = 1e18 + 7;
const ull mod = 9223372034707292160;
LL gcd(LL a, LL b) {return b ? gcd(b, a % b) : a;}
LL lcm(LL a, LL b) {return a / gcd(a, b) * b;}
LL powmod(LL a, LL b, LL MOD) {LL ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
double dpow(double a, LL b) {double ans = 1.0; while (b) {if (b % 2)ans = ans * a; a = a * a; b /= 2;} return ans;}
int main() {
#ifndef ONLINE_JUDGE
	FIN
#endif
	LL n, k;
	cin >> n >> k;
	if (k > n) {
		cout << "NO" << endl;
	} else {
		multiset<int> ans;
		multiset<int>::iterator it;
		for (int i = 0; i < 30; i++)
			if ((n >> i) & 1)
				ans.insert(i);
		if (ans.size() > k)
		{
			cout << "NO";
			return 0;
		}
		cout << "YES\n";
		while ((int)ans.size() < k)
		{
			it = ans.end();
			it--;
			int x = (*it);
			ans.erase(ans.lower_bound(x));
			ans.insert(x - 1);
			ans.insert(x - 1);
		}
		for (it = ans.begin(); it != ans.end(); it++)
			cout << (1 << *it) << " ";
		return 0;
	}
}

D. Circular Dance

传送门

题意：n个人围成一圈，每个人报出接下来两个人的序号，但是不保证按照顺序来，求解这一圈人的编号顺序，题目有spj

题解：将每个人报的编号想成两个点，然后就行成了一个图的关系，那么现在我们就只需要判定这个图的连通性即可

代码：

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define PI acos(-1)
#define eps 1e-8
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> PII;
const int maxn = 3e5 + 5;
const LL INF = 1e18 + 7;
const ull mod = 9223372034707292160;
LL gcd(LL a, LL b) {return b ? gcd(b, a % b) : a;}
LL lcm(LL a, LL b) {return a / gcd(a, b) * b;}
LL powmod(LL a, LL b, LL MOD) {LL ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
double dpow(double a, LL b) {double ans = 1.0; while (b) {if (b % 2)ans = ans * a; a = a * a; b /= 2;} return ans;}
vector<int> ans;
vector<int> mp[maxn];
bool check(int a, int b) {
	for (int i = 0; i < mp[a].size(); i++) {
		if (mp[a][i] == b) {
			return 1;
		}
	}
	return 0;
}
int get_next(int x){
	int v1=mp[x][0];
	int v2=mp[x][1];
	if(check(v1,v2)){
		return v1;
	}
	return v2;
}
int main() {
#ifndef ONLINE_JUDGE
	FIN
#endif
	int n;
	scanf("%d", &n);
	int u, v;
	for (int i = 1; i <= n; i++) {
		scanf("%d%d", &u, &v);
		mp[i].push_back(u);
		mp[i].push_back(v);
	}
	if (n == 3) {
		cout << "1 2 3" << endl;
		return 0;
	}
	ans.push_back(1);
	while (ans.size() < n) {
		int val = ans.back();
		int nxt = get_next(val);
		ans.push_back(nxt);
	}
	for (int i = 0; i < ans.size(); i++) {
		if (i)
			cout << " ";
		cout << ans[i];
	}
	cout << endl;
}

E. Almost Regular Bracket Sequence

传送门

题意：给你一个括号序列，你需要翻转其中一个括号使得括号序列合法，求应该翻哪个，有spj

题解：我们可以用前缀和来很好的解决括号匹配问题，首先我们规定‘（’是1，‘）’是-1求出这个序列的前缀和，然后用一个数组来记录从后往前的前缀和的最小值，最后从前往后扫一遍，判断翻转这个位置是否能够使得序列合法即可

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+5;
string str;
int n;
bool check(string str){
	int len=str.length();
	int tmp=len/2;
	for(int i=0;i<tmp;i++){
		if(str[i]!=str[len-i]) return 0;
	}
	return 1;
}
int sum[maxn];
int minn[maxn];
int main(){
	cin>>n>>str;
	sum[0]=0;
	for(int i=0;i<n;i++){
		if(str[i]=='(') sum[i+1]=sum[i]+1;
		else if(str[i]==')') sum[i+1]=sum[i]-1;
	}
	minn[n]=sum[n];
	for(int i=n-1;i>=0;i--){
		minn[i]=min(minn[i+1],sum[i]);
	}
	int ans=0;
	for(int i=0;i<n;i++){
		if(sum[i]<0) break;
		int tmp=sum[i];
		if(str[i]==')'){
			tmp++;
		}else if(str[i]=='('){
			tmp--;
		}
		if(tmp<0) continue;
		if(sum[n]-sum[i+1]+tmp!=0) continue;
		if(minn[i+1]-sum[i+1]+tmp<0) continue;
		ans++;
 	}
 	cout<<ans<<endl;
}

F. Make It Connected

传送门

题意：给你一个无向图，有n个点，每个点有一个权值，从a点走到b点的花费是a、b的权值和，有m条边可以连接，如果连接u和v则花费w的权值，当然也可以选择不连，求使得这个图联通的最小花费

题解：我们找到一个起点，要想使得这个生成这个图的花费最小，那么起点一定是权值最小的那个，连边时将这个起点和所有的点连接起来，然后最后跑一个最小生成树即可

代码

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define PI acos(-1)
#define eps 1e-8
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> PII;
const int maxn = 3e5 + 5;
const LL INF = 1e18 + 7;
const ull mod = 9223372034707292160;
LL gcd(LL a, LL b) {return b ? gcd(b, a % b) : a;}
LL lcm(LL a, LL b) {return a / gcd(a, b) * b;}
LL powmod(LL a, LL b, LL MOD) {LL ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
double dpow(double a, LL b) {double ans = 1.0; while (b) {if (b % 2)ans = ans * a; a = a * a; b /= 2;} return ans;}
int n, m;
LL a[maxn];
struct EDGE {
	int u, v;
	LL w;
	bool operator < (const EDGE&a) {
		return w < a.w;
	}
}edge[maxn<<2];
int f[maxn];
int find(int x){
	return x==f[x]?x:f[x]=find(f[x]);
}
int main() {
#ifndef ONLINE_JUDGE
	FIN
#endif
	cin>>n>>m;
	int minn=0;
	a[0]=INF;
	for(int i=1;i<=n;i++){
		scanf("%lld",&a[i]);
		if(a[i]<a[minn]) minn=i;
	}
	for(int i=1;i<=n;i++){
		edge[i]=(EDGE){i,minn,a[i]+a[minn]};
		f[i]=i;
	}
	for(int i=n+1;i<=n+m;i++){
		scanf("%d%d%lld",&edge[i].u,&edge[i].v,&edge[i].w);
	}
	sort(edge+1,edge+1+n+m);
	LL ans=0;
	for(int i=1;i<=n+m;i++){
		int u=find(edge[i].u);
		int v=find(edge[i].v);
		if(u!=v){
			f[u]=v;
			ans+=edge[i].w;
		}
	}
	cout<<ans<<endl;
}