# 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;
}

原文地址：https://www.cnblogs.com/buerdepepeqi/p/10198368.html

时间： 2024-10-14 07:57:38

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