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A - αlphabet

Problem Summary

Input an English letter $a$ and determine whether it is uppercase or lowercase.

Input Format

$a$

Output Format

If $a$ is lowercase, output a;
If $a$ is uppercase, output A.

Sample Input 1

1

B

Sample Output 1

1

A

Since B is uppercase, output A.

Sample Input 2

1

a

Sample Output 2

1

a

Since a is lowercase, output a.

Code

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#include <cstdio>
using namespace std;

int main(int argc, char** argv)
{
    char a = getchar();
    if('a' <= a && a <= 'z') putchar('a');
    else putchar('A');
    return 0;
}

B - Mix Juice

Problem Summary

Given an array $p_1, p_2, p_3, ..., p_N$ of length $N$, select $K$ elements such that their sum is minimized.

$1\le K\le N\le 1000$
$1\le p_i\le 1000$ ($1\le i\le N$)

Input Format

$N K$
$p_1~p_2~p_3~\dots~p_N$

Output Format

One line containing the minimal sum.

Sample Input 1

1
2


5 3
50 100 80 120 80

The minimal sum is $50+80+80=210$.

Sample Output 1

1

210

Since B is uppercase, output A.

Sample Input 2

Sample Output 2

1

1000

Only one element exists, so the minimal sum is $1000$. (Note added by author)

Code

To minimize the sum, select the $K$ smallest elements in the array. Use the sort() function from <algorithm> to sort the array in ascending order, then sum the first $K$ elements.

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#include <cstdio>
#include <algorithm>
#define maxn 1005
using namespace std;

int a[maxn];

int main(int argc, char** argv)
{
    int n, k, sum = 0;
    scanf("%d%d", &n, &k);
    for(int i=0; i<n; i++)
        scanf("%d", a + i);
    sort(a, a + n);
    for(int i=0; i<k; i++)
        sum += a[i];
    printf("%d\n", sum);
    return 0;
}

C - One Quadrillion and One Dalmatians

Problem Summary

There are $1000000000000001$ ($10^{15}+1$) dogs named as:
a, b, .., z, aa, ab, .., az, ba, bb, .., bz, .., za, zb, .., zz, aaa, aab, .., aaz, aba, abb, .., abz, …, zzz, aaaa, …
Question: What is the name of the $N$-th dog?

$1\le N\le 10^{15}+1$

Input Format

$N$

Output Format

One line containing the name of the $N$-th dog.

Samples

Multiple samples are consolidated into a table for brevity:

Input	Output
2	b
27	aa
123456789	jjddja

Code

This is equivalent to converting a decimal number to a base-26 representation:

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#include <cstdio>
#define BASE 26LL
using namespace std;

int main(int argc, char** argv)
{
    char s[12];
    int cnt = 0;
    long long n;
    scanf("%lld", &n);
    while(n > 0)
    {
        n --;
        s[cnt++] = n % BASE + 'a';
        n /= BASE;
    }
    for(int i=cnt-1; i>=0; i--) putchar(s[i]);
    putchar('\n');
    return 0;
}

D - Replacing

Problem Summary

Given an array $A_1, A_2, \dots, A_N$, perform $Q$ operations:

In the $i$-th operation, replace all occurrences of $B_i$ with $C_i$.
Output the sum of all elements in $A$ after each operation (denoted as $S_i$).

$1\le N, Q, A_i, B_i, C_i\le 10^5$ ($1\le i\le N$)
$B_i\ne C_i$

Input Format

$N$
$A_1~A_2~...~A_N$
$Q$
$B_1~C_1$
$B_2~C_2$
$:$
$B_Q~C_Q$

Output Format

$S_1$
$S_2$
$:$
$S_N$

Sample Input 1

Sample Output 1

Step	Array $A$
Initial	$\{1, 2, 3, 4\}$
$i=1$	$\{2, 2, 3, 4\}$
$i=2$	$\{2, 2, 4, 4\}$
$i=3$	$\{4, 4, 4, 4\}$

Sample Input 2

Note: $B_i$ may not exist in the array.

Sample Output 2

Sample Input 3

Sample Output 3

Code

Use an array to track the frequency of each value in $A$. Also maintain the total sum of the array.

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#include <cstdio>
#define maxn 100005
using namespace std;

typedef long long LL;

int cnt[maxn];
LL sum = 0;

inline LL s(const LL& i)
{
    return cnt[i] * i;
}

int main(int argc, char** argv)
{
    int n, q;
    scanf("%d", &n);
    for(int i=0; i<n; i++)
    {
        int t;
        scanf("%d", &t);
        sum += t;
        cnt[t] ++;
    }
    scanf("%d", &q);
    while(q--)
    {
        int x, y;
        scanf("%d%d", &x, &y);
        sum -= s(x) + s(y);
        cnt[y] += cnt[x];
        cnt[x] = 0;
        sum += s(y);
        printf("%lld\n", sum);
    }
    return 0;
}