A - Xenny and Alternating Tasks

Problem surface

Title Solution

Enumerate who did it on the first day and take the two answers to (min).

Code

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)

using namespace std;

inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}

int t, n, a[20003], b[20003];

int main()
{
    //File("XENTASK");
    t = gi();
    while (t--)
    {
        n = gi();
        for (int i = 1; i <= n; i+=1) a[i] = gi();
        for (itn i = 1; i <= n; i+=1) b[i] = gi();
        itn ans = 0, sum = 0;
        for (itn j = 1; j <= n; j+=1)
        {
            if (j & 1) sum = sum + a[j];
            else sum = sum + b[j];
        }
        for (int j = 1; j <= n; j+=1)
        {
            if (j & 1) ans = ans + b[j];
            else ans = ans + a[j];
        }
        printf("%d\n", min(ans, sum));
    }
    return 0;
}

B - Bear and Extra Number

Problem surface

Title Solution

Sort the arrays, traverse the elements of the arrays, and then classify and discuss:

1. If \(a_i = a {i+1}), then output (a_i).
2. If u (a {i+1}-a_i>1):
   • Because there is a unique solution to the title, so (i) can only be (n-1) or (1\).
     • If \(i) is (n-1), then output (a_n\).
     • If (i) is (1), then output (a_1).

Code

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <cctype>
#define gI gi
#define itn int
#define File(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout)

using namespace std;

inline int gi()
{
    int f = 1, x = 0; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {x = x * 10 + c - '0'; c = getchar();}
    return f * x;
}

int t, n, a[100003], ans, sum;

int main()
{
    //File("EXTRAN");
    t = gi();
    while (t--)
    {
        n = gi();
        for (itn i = 1; i <= n; i+=1) a[i] = gi();
        sort(a + 1, a + 1 + n);
        for (int i = 1; i < n; i+=1)
        {
            if (a[i + 1] == a[i]) {printf("%d\n", a[i]); break;}
            if (a[i + 1] - a[i] > 1)
            {
                if (i == n - 1) {printf("%d\n", a[n]); break;}
                else if (i == 1) {printf("%d\n", a[1]); break;}
            }
        }
    }
    return 0;
}

C - Cooking Schedule

Problem surface

Title Solution

Two points.

The key is how to write (check).

There's no time today. First dig a hole and then fill it tomorrow.