Codeforces Round #666 (Div. 2)题解ABC

A. Juggling Letters

题目传送门

Juggling Letters

题目大意

给你n个字符串，让你移动任意次后判断能否让每个字符串相等

思路

直接统计每个字符串中出现的字符串的个数和，判断和是否为n的倍数

AC Code

#include<cstdio>#include<algorithm>#include<iostream>#include<cstring>using namespace std;#define endl "\n"#define INF 0x3f3f3f3f#define int long long// #define TDS_ACM_LOCALconst int N=1009;int n, vis[30];string s[N];void solve(){cin>>n;memset(vis, 0, sizeof(vis));for(int i=1; i<=n; i++){cin>>s[i];for(int j=0; j<s[i].length(); j++){vis[s[i][j]-"a"]++;}}int flag=0;for(int i=0; i<27; i++){if(vis[i]!=0 && vis[i]%n!=0) flag=1;}if(flag)cout<<"NO"<<endl;elsecout<<"YES"<<endl;return ;}signed main(){ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);#ifdef TDS_ACM_LOCALfreopen("D:\\VS code\\.vscode\\testall\\in.txt", "r", stdin);freopen("D:\\VS code\\.vscode\\testall\\out.txt", "w", stdout);#endifint T;cin>>T;while(T--)solve();return 0;}

B. Power Sequence

题目传送门

Power Sequence

题目大意

给你一个长度为n的数组a，下标从$0$0到$n-1$n−1，你可以重新排列数组a的顺序，然后对数组中的任意一个元素+1或者-1，其花费都是1，求最小花费，使得$a_i=c_i$ai​=ci​

思路

枚举C的取值，当代价开始增加时退出，增加前的代价即为最小代价

AC Code

#include<cstdio>#include<algorithm>#include<iostream>#include<cstring>#include<cmath>using namespace std;#define endl "\n"#define INF 0x3f3f3f3f3f3f3f#define int long long// #define TDS_ACM_LOCALconst int N=1e5 +9;int n, a[N];void solve(){cin>>n;for(int i=1; i<=n; i++) cin>>a[i];sort(a+1, a+1+n);int ans=INF, tep;for(int c=1; 1; c++){tep=a[1]-1;int mj=c;for(int j=2; j<=n; j++){tep+=abs(a[j]-mj);if(tep>ans)break;mj*=c;}if(tep>ans)break;ans=tep;}cout<<ans<<endl;return ;}signed main(){ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);#ifdef TDS_ACM_LOCALfreopen("D:\\VS code\\.vscode\\testall\\in.txt", "r", stdin);freopen("D:\\VS code\\.vscode\\testall\\out.txt", "w", stdout);#endifsolve();return 0;}

C. Multiples of Length

题目传送门

C. Multiples of Length

题目大意

给你一个长度为n的数组，需要使用三步将数组中的每个元素变为0
每步的操作为，选择一个区间，对区间中的值进行加减k的操作，但是k必须为区间的长度的倍数

思路

当n=1的时候很简单，直接第一步减到0即可，后两次操作0
不为0时
很明显不会让你一步就完成，所以可以操作前n-1个
可以考虑第一步操作前n-1个数使其变成整个数组的倍数
第二步操作第n个数使其变成0
第三步选择整个区间变成0

对前面n-1个数操作为$a_i+k(n-1)$ai​+k(n−1)，即为$kn+a_i-k$kn+ai​−k，所以可以直接k取$a_i$ai​，使得每位都为n的倍数

AC Code

#include<cstdio>#include<algorithm>#include<iostream>#include<cstring>using namespace std;#define endl "\n"#define INF 0x3f3f3f3f#define int long long// #define TDS_ACM_LOCALconst int N=1e5 +9;int n, a[N];void solve(){cin>>n;for(int i=1; i<=n; i++) cin>>a[i];if(n==1){cout<<"1 1"<<endl;cout<<-a[1]<<endl;for(int i=2; i<=3; i++){cout<<"1 1"<<endl;cout<<"0"<<endl;}return ;}cout<<"1 "<<n-1<<endl;for(int i=1; i<=n-1; i++){cout<<a[i]*(n-1)<<" ";a[i]+=(n-1)*a[i];}cout<<endl;cout<<n<<" "<<n<<endl;cout<<-a[n]<<endl;cout<<"1 "<<n<<endl;for(int i=1; i<=n-1; i++){cout<<-a[i]<<" ";}cout<<"0"<<endl;return ;}signed main(){ios::sync_with_stdio(0);cin.tie(0), cout.tie(0);#ifdef TDS_ACM_LOCALfreopen("D:\\VS code\\.vscode\\testall\\in.txt", "r", stdin);freopen("D:\\VS code\\.vscode\\testall\\out.txt", "w", stdout);#endifsolve();return 0;}