#总结# 5 - 第三周周五/3h-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_45797295/article/details/117373672

本文涵盖了多项编程竞赛的题目解析，包括团队配合判断、数值计算、青蛙问题及矩阵检查等。A-ASecret主要考察团队成员间的关系，通过矩阵判断是否为‘BadTeam!’。D-Minimum’sRevenge解决最小复仇问题，E-Mr.Frog’sProblem涉及青蛙跳跃的最短路径。F-Mr.Frog’sGame检查棋盘是否存在相同元素的连续路径。每个问题都提供了AC代码和思路总结。

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A - A Secret

题目

思路

代码

总结

B - Friend-Graph

#include<iostream>
#define maxx 30
using namespace std;
typedef long long ll;
int a[maxx][maxx];
int main()
{
    int T;
    cin>>T;
    while (T--)
    {
        int n;
        cin>>n;
        if (n>6)
        {
            cout<<"Bad Team!"<<endl;
            break;
        }

        for (int i = 1; i < n; ++i)
        {
            for (int j = i+1; j <= n; ++j)
            {
                cin>>a[i][j];
                a[j][i] = a[i][j];
            }
        }
//        for (int i = 1; i <= n; ++i)
//        {
//            for (int j = 1; j <= n; ++j)
//            {
//                cout<<a[i][j]<<' ';
//            }
//            cout<<endl;
//        }
        int f=0;
        for (int i = 1; i <= n; ++i)
        {
            int c1=0,c2=0;
            for (int j = 1; j <= n; ++j)
            {
                if (a[i][j]==0)
                    c1++;
                else
                    c2++;
                if (c1>=4 || c2>=3)
                {
                    f=1;
                    break;
                }
            }
            c1=0,c2=0;
            for (int j = 1; j <= n; ++j)
            {
                if (a[j][i]==0)
                    c1++;
                else
                    c2++;
                if (c1>=4 || c2>=3)
                {
                    f=1;
                    break;
                }
            }
            if (f==1)
                break;
        }
        if (f==1)
            cout<<"Bad Team!"<<endl;
        else
            cout<<"Great Team!"<<endl;
    }
    return 0;
}

D - Minimum’s Revenge

#include<iostream>
#define maxx 30
using namespace std;
typedef long long ll;

int main()
{
    int n;
    cin>>n;
    for (int i = 1; i <= n; ++i)
    {
        ll x;
        cin>>x;
        printf("Case #%d: ",i);
        cout<<(1+x)*x/2-1<<endl;
    }
    return 0;
}

E - Mr. Frog’s Problem

ac
猜的，一开始忘了ll格式

#include<iostream>

#define maxx 30
using namespace std;
typedef long long ll;

int main()
{
    int t;
    cin >> t;
    ll cnt = 0;
    while (t--)
    {
       ll a, b;
        cin >> a >> b;
        printf("Case #%d:\n", ++cnt);
        if (a == b)
        {
            cout << 1 << endl;
            cout << a << ' ' << b<< endl;
        }
        else
        {
            cout << 2 << endl;
            if (a < b)
            {
                cout << a << ' ' << b<< endl;
                cout << b << ' ' << a<< endl;
            }
            else
            {
                cout << b << ' ' << a<< endl;
                cout << a << ' ' << b<< endl;
            }
        }
    }
    return 0;
}

F - Mr. Frog’s Game

ac
要注意的是遍历从一开始边边上：不仅是对[1][1][n][m]的，而是一列中有相同的即可，然后在中间是必须要相邻才阔以~

#include<iostream>

#define maxx 50
using namespace std;
typedef long long ll;
int a[maxx][maxx];
int main()
{
    int t;
    cin>>t;
    for (int k = 1; k <= t; ++k)
    {
        int n,m;
        cin>>n>>m;
        int f = 0;
        for (int i = 1; i <= n; ++i)
        {
            for (int j = 1; j <= m; ++j)
            {
                cin>>a[i][j];
            }
        }
        //bianbian
        for (int i = 1; i < m; ++i)
        {
            for (int j = i+1; j <= m; ++j)
            {
                if (a[1][i] == a[1][j] || a[n][i]==a[n][j])
                    f=1;
            }
        }
        for (int i = 1; i < n; ++i)
        {
            for (int j = i+1; j <= n; ++j)
            {
                if (a[i][1] == a[j][1] || a[i][m]==a[j][m])
                    f=1;
            }
        }
        //zhongjian
        for (int i = 1; i <= n; ++i)
        {
            for (int j = 1; j <= m; ++j)
            {
                if (i!=n &&a[i][j] == a[i+1][j])
                    f=1;
                if (j!=m && a[i][j]==a[i][j+1])
                    f=1;
            }
        }
        printf("Case #%d: ",k);
        if(f==1)
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    }
    return 0;
}