Div2第一次AC兩道題

A題：
直接比較一張卡片上的兩個數，blue大blue++,red大red++，最後比較（當時就憑感覺應該是這樣qwq）

#include<iostream>
#include<stdio.h>
#include<string>
#include<algorithm>
using namespace std;

char r[1005];
char b[1005];

int main()
{
    int t;
    cin>>t;
    int n;
    int blue;
    int red;
    
 
while(t--){
        cin>>n;
        red = blue = 0;
        for (int i = 1;i<=n;i++){
            cin >> r[i];
        }
        for (int i = 1; i <= n;i++){
            cin >> b[i];
        }
        for (int i = 1; i <= n;i++){
            if((int)(r[i]-'0')>(int)(b[i]-'
 
0'))
                red++;
            if((int)(r[i]-'0')<(int)(b[i]-'0'))
                blue++;
        }
        if (red > blue)
            cout<<"RED"<<endl;
        if(red<blue)
            cout << "BLUE" << endl;
        if(red==blue)
            cout << "
 
EQUAL" << endl;
    }
    return 0;
}

B題：

一開始用dfs搜，明顯超時，果斷用dfs找規律

#include<iostream>
using namespace std;
//dfs打表wow
//前面是dfs搜的程式碼
/*
void dfs(int x,int y)
{
    if(num==n){
        if(!vis[x][y]){
            vis[x][y] = 1;
            sum++;
        }
        return;
    }
    for (int i = 1;i<=2;i++){
        if (flag)
        {
            num++;
            flag = 0;
            dfs(x + 1, y);
            num--;
            num++;
            flag = 0;
            dfs(x - 1, y);
            num--;
            return;
        }
        else
        {
            num++;
            flag = 1;
            dfs(x, y + 1);
            num--;
            num++;
            flag = 1;
            dfs(x, y - 1);
            num--;
            return;
        }
    }
}



int main()
{
    cin>>n;
    flag = 0;
    dfs(500,500);
    long long k=sum;
    cout << sum << endl;
    flag = 1;
    dfs(500, 500);
    cout<<sum-k<<endl;
    cout << sum;
    return 0;
}
*/

int main()
{
    int n;
    cin >> n;
    if(n%2==0){
        cout << (n / 2 + 1) * (n / 2 + 1);
        return 0;
    }else{
        n++;
        cout << ((n / 2 + 1) * (n / 2 + 1) - (n / 2 + 1))*2;
        return 0;
    }
}

C題：（補題）

比賽時沒做出來（其實是不會）

（數論一直搞不懂）

gcd(a,b)=gcd(a-b,b)

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;


long long gcd(long long x,long long y)
{
    if(y==0)
        return x;
    else
        return gcd(y, x % y);
}
long long a[200006];

int main()
{
    int n;
    int m;
    long long t;
    long long gcd_x=1;
    cin >> n >> m;
    for(int i=1;i<=n;i++)
        cin>>a[i];
    sort(a + 1, a + n + 1);
    if(n>=2){
        gcd_x = a[2] - a[1];
        for (int i = 3; i <= n;i++)
            gcd_x=gcd(gcd_x,a[i]-a[i-1]);
        for (int i = 1;i<=m;i++){
            cin>>t;
            cout << gcd(gcd_x, a[1] + t) << ' ';
        }
    }else{
        for (int i = 1; i <= m;i++){
            cin >> t;
            cout << t + a[1] << ' ';
        }
    }
    return 0;
}