Codeforces Round #634 (Div. 3)

A

题目链接：http://codeforces.com/contest/1335

思路：简单的推理：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
            ll a;
            scanf("%lld",&a);
            if(a<=2)
                printf("0\n");
            else if(a%2==0)
                printf("%lld\n",a/2-1);
            else
                printf("%lld\n",a/2);
    }

    return 0;
}

　　B

思路：逐位考虑，保持循环即可，字符串拥有循环节

//-------------------------------------------------
//Created by HanJinyu
//Created Time :三  4/15 23:10:31 2020
//File Name :cf3B.cpp
//-------------------------------------------------

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
#define long long ll
#define double db
#define maxn 200005
using namespace std;
void disp(int a,int b)
{
    for(int i=1;i<=a/b;i++)
    {
        for(int j=0;j<b;j++)
        {
            printf("%c",97+j);
        }
    }
    for(int i=0;i<a%b;i++)
        printf("%c",97+i);

}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,a,b;
        scanf("%d%d%d",&n,&a,&b);
        int num=n/a;
        int num1=n%a;
        for(int k=0;k<num;k++)
            disp(a,b);
        int ops=0;
        for(int t=a%b;t<a%b+num1;t++)
        {
            if(t>=b)
            {
                break;
            }
            else{
                 ops++;
                 printf("%c",97+t);
            }

        }
        if(num1-ops<=b){

        for(int i=0;i<(num1-ops);i++)
            printf("%c",97+i);

        }
        else
            disp(num1-ops,b);
        printf("\n");
    }
     return 0;
}

　　c

思路：比较数字出现的最多次数那个数字的次数和数组不同的个数，输出最小的即可

//
// Created by H on 2020/4/13.
//

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        map<int,int>ap;
        int n;
        scanf("%d",&n);
        int mm;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&mm);
            ap[mm]++;
        }
        int cishu=0;
        for(auto v:ap)
        {
            cishu=max(cishu,v.second);
        }
        if(ap.size()==cishu)
            cout<<cishu-1<<endl;
        if(ap.size()>cishu)
            cout<<cishu<<endl;
        if(ap.size()<cishu)
            cout<<ap.size()<<endl;
    }
    return 0;
}

　　D

思路：数独，只要任意将一个数变成另外一个数即可

//-------------------------------------------------
//Created by HanJinyu
//Created Time :四  4/16 13:35:36 2020
//File Name :shudu.cpp
//-------------------------------------------------

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
#define long long ll
#define double db
#define maxn 200005
using namespace std;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int t;
    scanf("%d",&t);
    while(t--)
    {
        char ch[10][10];
        for(int i=0;i<9;i++)
        {

            for(int j=0;j<9;j++)
                {

                    cin>>ch[i][j];
                    if(ch[i][j]=='1')
                        ch[i][j]='2';
                }
        }
        for(int i=0;i<9;i++)
        {

            for(int j=0;j<9;j++)
            {

                cout<<ch[i][j];
            }
            cout<<endl;
        }
    }
 
     return 0;
}