# Codeforces Round #529(Div.3)個人題解
阿新 • • 發佈：2018-12-30
Codeforces Round #529(Div.3)個人題解
前言： 閒來無事補了前天的cf，想著最近刷題有點點怠惰，就直接一場cf一場cf的刷算了，以後的題解也都會以每場的形式寫出來
A. Repeating Cipher
題意：第一個字母寫一次，第二個字母寫兩次，依次遞推，求原字串是什麼
題解：1、2、3、4，非常明顯的d=1的等差數列，所以預處理一個等差數列直接取等差數列的每一項即可
程式碼：
#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
題意：給你一串數字，要你刪除一個數最小化這串數字中最大值-最小值的差
題解：用multiset存一下，然後討論刪去最大的數更好還是刪去最小的數更好
程式碼：
#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
題意：給你一個數n，要求你用k個2的冪次數去拼出這個數，如果不能輸出-1
題解：先將n轉換為相對應的二進位制數 ，如果n的二進位制數中的1的個數大於k，顯然是沒有解的，如果n的二進位制數中的1的個數小於1，那麼就把每一個大於2的數分解（x->x/2+x/2），湊出k個1即可，最後輸出 一下就行
程式碼
#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
題意：n個人圍成一圈，每個人報出接下來兩個人的序號，但是不保證按照順序來，求解這一圈人的編號順序，題目有spj
題解：將每個人報的編號想成兩個點，然後就行成了一個圖的關係，那麼現在我們就只需要判定這個圖的連通性即可
程式碼：
#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
題意：給你一個括號序列，你需要翻轉其中一個括號使得括號序列合法，求應該翻哪個，有spj
題解：我們可以用字首和來很好的解決括號匹配問題，首先我們規定‘（’是1，‘）’是-1求出這個序列的字首和，然後用一個數組來記錄從後往前的字首和的最小值，最後從前往後掃一遍，判斷翻轉這個位置是否能夠使得序列合法即可
程式碼
#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
題意：給你一個無向圖，有n個點，每個點有一個權值，從a點走到b點的花費是a、b的權值和，有m條邊可以連線，如果連線u和v則花費w的權值，當然也可以選擇不連，求使得這個圖聯通的最小花費
題解：我們找到一個起點，要想使得這個生成這個圖的花費最小，那麼起點一定是權值最小的那個，連邊時將這個起點和所有的點連線起來，然後最後跑一個最小生成樹即可
程式碼
#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;
}