标签：ns2 -- force 题解 cto tmp void link racket
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
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
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
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
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
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
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;
}# Codeforces Round #529(Div.3)个人题解
标签：ns2 -- force 题解 cto tmp void link racket
原文地址：https://www.cnblogs.com/buerdepepeqi/p/10198368.html
