Educational Codeforces Round 49 (Rated for Div. 2)

 Educational Codeforces Round 49 (Rated for Div. 2)

题目链接：https://codeforces.com/contest/1027

A题题解

  • 题意：根据一个字符串，经过变化后是否能成为一个回文串（变成相邻的字母，a只能变成b，z只能变成x，其他的有两种可能）（都要变）
  • 很简单，只要不满足两个条件比较ASCII码1.相差不为1  2.相差大于2就不行，其他就行
  • AC代码
  • #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
//ll a ,b ,c ,d;
char a[200];
const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18
int main(){
    int n;
    cin >> n;
    while(n --){
        int t;
        cin >> t;
        memset(a,0,sizeof(a));
        cin >> a;
        bool flag = 1;
        if(t % 2 == 0){
            for(int i = 0 ; i < t; i++){
                if(a[i] != a[t-i-1]){
                    if(abs(a[i] - a[t-i-1] ) > 2 || abs(a[i] - a[t-i-1] ) == 1 )
                    {
                        flag = 0;
                        break;
                    }
                    //cout << "NO" << endl;
                    
                
                }
            }
        }
        if(!flag) cout << "NO" << endl;
        else cout << "YES" << endl;
    }
}#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
//ll a ,b ,c ,d;
char a[200];
const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18
int main(){
    int n;
    cin >> n;
    while(n --){
        int t;
        cin >> t;
        memset(a,0,sizeof(a));
        cin >> a;
        bool flag = 1;
        if(t % 2 == 0){
            for(int i = 0 ; i < t; i++){
                if(a[i] != a[t-i-1]){
                    if(abs(a[i] - a[t-i-1] ) > 2 || abs(a[i] - a[t-i-1] ) == 1 )
                    {
                        flag = 0;
                        break;
                    }
                    //cout << "NO" << endl;
                    
                
                }
            }
        }
        if(!flag) cout << "NO" << endl;
        else cout << "YES" << endl;
    }
}
    View Code

     

B题题解

  • 题意：在矩阵里面放数，先把横纵坐标之和是偶数的放满，从左往右，从上往下，然后从最小的奇数（横纵坐标之和）开始继续往后放数；直到把n*n这些数放完
  • 数学推公式，ans1 = n * (a - 1) + b + 1; a+b为奇数：ans = ans1 + n * n ; 为偶数 ：ans1 ; 然后都 / 2；
  • 可以分类讨论，讨论可以发现共性
  • AC代码
  • #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
//ll a ,b ,c ,d;
//char a[200];
const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18
int main(){
ll n , t;
cin >> n >> t;
ll ans ;
while(t--){
    ll a , b;
    cin >> a >> b;
    ans = 0;

ans = n * (a - 1) + b + 1;
if((a + b)%2!=0)
ans += n * n;
     cout << ans / 2<< endl;
}
return 0;
}
    View Code

 

C题题解

  • AC代码
  • #include <cstdio>
#include <cctype>
#include <map>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
inline char in() {
    static char buf[10001],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,10000,stdin),p1==p2)?EOF:*p1++;
}
template <class Tp> void read(register Tp &s) {
    s=0;
    register bool neg=0;
    register char c;
    while(!isdigit(c=in())) if(c=='-') neg=1;
    while(s=(s<<3)+(s<<1)+c-48,isdigit(c=in()));
    s=(neg?-s:s);
}

int T,n,a[1000005];
std::map<int,int> cnt;
int main() {
    for(read(T);T;--T) {
        cnt.clear();
        read(n);
        int t,m=0;
        FOR(i,1,n) read(t),cnt[t]++;
        bool flg=0;
        for(std::map<int,int>::iterator it=cnt.begin();it!=cnt.end();++it) {
            if((*it).second>=2) a[++m]=(*it).first;
            if((*it).second>=4) {
                flg=1;
                int ans=(*it).first;
                printf("%d %d %d %d\n",ans,ans,ans,ans);
                break;
            }
        }
        if(flg) continue;
        int x=a[1],y=a[2];
        FOR(i,3,m) if(x*a[i]<a[i-1]*y) x=a[i-1],y=a[i];
        printf("%d %d %d %d\n",x,x,y,y);
    }
    return 0;
}
    View Code

     

 D题题解

  • 题意：一个寝室里有n个房间和一个老鼠，老鼠一开始可能在任意一个房间，老鼠会从房间i跳到a[i]房间，问在哪些房间下陷阱可以用最小的代价抓到老鼠？
  • 分析：遍历所有房间，给从这些可能到达的房间打上标记，每次打标志在花费最小的房间布下陷阱
  • AC代码 
    #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
%:pragma GCC optimize("Ofast")
%:pragma GCC optimize("inline")
%:pragma GCC optimize("-fgcse")
%:pragma GCC optimize("-fgcse-lm")
%:pragma GCC optimize("-fipa-sra")
%:pragma GCC optimize("-ftree-pre")
%:pragma GCC optimize("-ftree-vrp")
%:pragma GCC optimize("-fpeephole2")
%:pragma GCC optimize("-ffast-math")
%:pragma GCC optimize("-fsched-spec")
%:pragma GCC optimize("unroll-loops")
%:pragma GCC optimize("-falign-jumps")
%:pragma GCC optimize("-falign-loops")
%:pragma GCC optimize("-falign-labels")
%:pragma GCC optimize("-fdevirtualize")
%:pragma GCC optimize("-fcaller-saves")
%:pragma GCC optimize("-fcrossjumping")
%:pragma GCC optimize("-fthread-jumps")
%:pragma GCC optimize("-funroll-loops")
%:pragma GCC optimize("-fwhole-program")
%:pragma GCC optimize("-freorder-blocks")
%:pragma GCC optimize("-fschedule-insns")
%:pragma GCC optimize("inline-functions")
%:pragma GCC optimize("-ftree-tail-merge")
%:pragma GCC optimize("-fschedule-insns2")
%:pragma GCC optimize("-fstrict-aliasing")
%:pragma GCC optimize("-fstrict-overflow")
%:pragma GCC optimize("-falign-functions")
%:pragma GCC optimize("-fcse-skip-blocks")
%:pragma GCC optimize("-fcse-follow-jumps")
%:pragma GCC optimize("-fsched-interblock")
%:pragma GCC optimize("-fpartial-inlining")
%:pragma GCC optimize("no-stack-protector")
%:pragma GCC optimize("-freorder-functions")
%:pragma GCC optimize("-findirect-inlining")
%:pragma GCC optimize("-fhoist-adjacent-loads")
%:pragma GCC optimize("-frerun-cse-after-loop")
%:pragma GCC optimize("inline-small-functions")
%:pragma GCC optimize("-finline-small-functions")
%:pragma GCC optimize("-ftree-switch-conversion")
%:pragma GCC optimize("-foptimize-sibling-calls")
%:pragma GCC optimize("-fexpensive-optimizations")
%:pragma GCC optimize("-funsafe-loop-optimizations")
%:pragma GCC optimize("inline-functions-called-once")
%:pragma GCC optimize("-fdelete-null-pointer-checks")
# define IOS ios::sync_with_stdio(false); cin.tie(0);cout.tie(0)
ll a[200000+ 10] , c[200000 + 10];
ll vis[200000 + 10];//记录是否走过 
int main(){
    IOS;
    ll n;
    ll sum=0 , v;
    cin >> n;
    for(int i = 1 ; i <= n ; i ++){
        cin >> a[i];
    }
    for(int i = 1 ; i <= n ; i ++){
        cin >> c[i];
    }
    ll x;
    for(int i = 1 ;i <= n ; i++){
    x = i;
    while( !vis[x]){
        vis[x] = i;
        x = c[x];
    }
    
    //当有重复时，即已经走过，就没必要再走 
    if(i != vis[x]) continue ;
     v = x;
    ll ans = a[x]; 
    while(v != c[x]) {
        x = c[x];
        ans = min(a[x] ,ans);
    }
//    cout << ans << endl;
    sum += ans ;
}
cout << sum << endl;
}
    View Code

     

顺带提一下，用加速挂真的快了很多，足足快了4倍（可以用scanf，还是用吧），不过强制氧化没什么用，只快1ms，应该是cf已经开了氧化

posted @ 2019-03-27 01:36  Agnel_Cynthia  阅读(249)  评论(0)  编辑  收藏  举报
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