AISing Programming Contest 2021(AtCoder Beginner Contest 202)(A-C)D待补

简介: 算法

链接:


A - Three Dice

题意:给你三个整数分别对应其骰子的大小,在这里,每一个骰子都是一个标准的立方骰子,其中相对面上的数字之和是7,求对面的和.

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int a,b,c;
    cin>>a>>b>>c;
    cout<<abs(7-a)+abs(7-b)+abs(7-c)<<endl;
}


B - 180°


题意:将字符串反转,并且把6换成9,把9换成6.

#include<bits/stdc++.h>
using namespace std;
int main()
{
    string s1;
    cin>>s1;
    reverse(s1.begin(),s1.end());
    for(int i=0;i<s1.length();i++){
        if(s1[i]=='9') cout<<6;
         else if(s1[i]=='6')    cout<<9;
         else cout<<s1[i];
    }
    cout<<endl;
}


C - Made Up


题意:

1.png

#include<bits/stdc++.h>
using namespace std;
#define int long long
const int maxn=1e5+100;
int a[maxn];
int b[maxn];
int c1[maxn];
signed  main()
{
    int n,m,i,j;
    int ans=0;
    cin>>n;
    map<int ,int >ma,mp;
    for(i=0;i<n;i++){
        cin>>a[i];
        ma[a[i]]++;
    }
    for(i=0;i<n;i++){
        cin>>b[i];
        mp[i]=b[i];
    }
    for(i=0;i<n;i++){
        cin>>m;
        c1[i]=mp[m-1];
        ans+=(ma[mp[m-1]]);
    }
    cout<<ans<<endl;
}


D - aab aba baa


题意:给你a个字符‘a’和b个字符’b’,输出按照字典序排列后第k大的

思路:

听说是dp,但是我没看太懂,再想一想,dp太弱鸟~

2.png

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