A - Palindromic Number

题意：输入三位数判断是否为回文串

#include<cstdio>
#include<iostream>
using namespace std;
string S;
int main()
{
    cin>>S;
    cout<<(S[0] == S[2] ? "Yes" : "No")<<endl;
    return 0;
}

B - Two Switches

题意：AB两人在操作同一个机器人，A在a时刻按下按钮持续b秒，B在c时刻按下按钮持续d秒，求两人时间段交集部分

#include<cstdio>
#include<iostream>
using namespace std;
int a,b,c,d;
int main()
{
    cin>>a>>b>>c>>d;
    cout<<(min(b,d) - max(a,c) > 0 ? min(b,d) - max(a,c) : 0)<<endl;
    return 0;
}

C - Meal Delivery

题意：x到a的绝对值距离与x到b的绝对值距离比较

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
int x,a,b;
int main()
{
    cin>>x>>a>>b;
    cout<<(abs(x - a) > abs(x - b) ? "B" : "A" )<<endl;
    return 0;
}

D - Not Found

题意：给出一个字符串，求字符串按字典序判断是否存在未出现的小写字母

#include<cstdio>
#include<iostream>
using namespace std;
string S;
char a[27] = {0};
int main()
{
    cin>>S;
    for (int i = 0; i < S.size(); i++)
        a[S[i] - 'a'] = 1;
    for (int i = 0; i < 26; i++)
        if(!a[i])
        {
            printf("%c\n",i+'a');
            return 0;
        }
    cout<<"None"<<endl;
    return 0;
}

E - Sandglass2

题意：沙漏⏳上半部分有x克沙子，每秒流走1克沙子，求t秒后剩下多少克沙子

#include<iostream>
#include<cstdio>
using namespace std;
int x,t;
int main()
{
    cin>>x>>t;
    cout<<(x - t >= 0 ? x - t : 0)<<endl;
    return 0;
}

F - Together

题意：给一个序列，每次可对序列元素进行+1、-1、维持原样其中一种操作，求经过操作后序列出现次数的元素是多少次

#include<cstdio>
#include<iostream>
#include<algorithm>
#define Maxn 100010
using namespace std;
int n,x,a[Maxn] = {0};
int main()
{
    cin>>n;
    for (int i = 0; i < n; i++)
    {
        cin>>x;
        a[x]++;
        a[x+1]++;
        a[x-1]++;
    }
    sort(a,a+Maxn);
    cout<<a[Maxn-1]<<endl;
    return 0;
}

G - September 9

题意：求输入的十进制两位数是否含有9

#include<iostream>
#include<cstdio>
using namespace std;
string S;
int main()
{
    cin>>S;
    cout<<( (S[0] == '9' || S[1] == '9') ? "Yes" : "No" )<<endl;
    return 0;
}

H - Generate Login

题意：给出用户名first name和last name，然后first name取前序和last name取前序构成的字符串字典序最小

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
string S,S2,t = "zzzzzzzzzz";
int main()
{
    cin>>S>>S2;
    for (int i = 0; i < S.size(); i++)
        for (int j = 0; j < S2.size(); j++)
            t = min(t,S.substr(0,i + 1) + S2.substr(0,j + 1));
    cout<<t<<endl;
    return 0;
}

I - Garden

题意：有一个长度为k的花园和n个大小不一的桶，求能否只用一个桶把整个花园的花浇完，至少要多少个小时

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int n,k,x,ans = 999;
int main()
{
    cin>>n>>k;
    for (int i = 0; i < n; i++)
    {
        cin>>x;
        if(k % x == 0)
            ans = min(ans,k / x);
    }
    cout<<ans<<endl;
    return 0;
}

J - Tricky Alchemy

题意：有A块黄水晶B块蓝水晶，已知制作黄球需要2块黄水晶，绿球需要1块黄水晶和1块蓝水晶，蓝球需要3块蓝水晶，求需要额外多少块水晶来制作

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
long long yellow,blue,a,b,x,y,z;
int main()
{
    cin>>yellow>>blue;
    cin>>x>>y>>z;
    a = x * 2 + y;
    b = y + 3 * z;
    cout<< max(0LL,a - yellow) + max(0LL,b - blue)<<endl;
    return 0;
}