LevOJ.sln - 第二期

LevOJ平台.sln

P1018 内卷矩阵

问题描述

解决方法

#include <stdio.h>

void fillMatrix(int n)
{
    int matrix[n][n];
    int top = 0, bottom = n - 1, left = 0, right = n - 1;
    int t = 1;

    while (t <= n * n)
    {
        for (int i = left; i <= right; i++)
        {
            matrix[top][i] = t++;
        }
        top++;

        for (int i = top; i <= bottom; i++)
        {
            matrix[i][right] = t++;
        }
        right--;

        if (top <= bottom)
        {
            for (int i = right; i >= left; i--)
            {
                matrix[bottom][i] = t++;
            }
            bottom--;
        }

        if (left <= right)
        {
            for (int i = bottom; i >= top; i--)
            {
                matrix[i][left] = t++;
            }
            left++;
        }
    }

    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            printf("%d ", matrix[i][j]);
        }
        printf("\n");
    }
}

int main()
{
    int n;
    while (scanf("%d", &n) != EOF)
    {
        fillMatrix(n);
    }

    return 0;
}

关于VLA变长数组

据说变长数组最开始是个错误

而且容易出安全漏洞 所以理论上不推荐使用

具体看文

P1020 最长最短单词

问题描述

解决方法

法一

#include <stdio.h>

int main()
{
    char str[500];
    fgets(str, 500, stdin);
    int p = 0, pmax, pmin, lmax = 0, lmin = 20;
    for (int i = 0; str[i] != '\0'; i++)
        if (str[i] == ' ')
        {
            if (i - p < lmin)
            {
                lmin = i - p;
                pmin = p;
            }
            if (i - p > lmax)
            {
                lmax = i - p;
                pmax = p;
            }
            p = i + 1;
        }

    for (int t = 0; t < lmax; t++)
        printf("%c", str[pmax + t]);
    printf("\0\n");
    for (int t = 0; t < lmin; t++)
        printf("%c", str[pmin + t]);
    printf("\0\n");

    return 0;
}

法二

#include <stdio.h>
#include <string.h>

int main()
{
    char str[500][500];
    size_t max = 0;
    size_t min = 500;
    char y[100][500], z[100][500];
    int i = 0;
    while (1)
    {
        scanf("%s", str[i]);
        i++;
        if (getchar() == '\n')
            break;
    }
    for (int j = 0; j < i; j++)
    {

        if (strlen(str[j]) > max)
        {
            max = strlen(str[j]);
            strcpy(y[0], str[j]);
        }
        if (strlen(str[j]) < min)
        {
            min = strlen(str[j]);
            strcpy(z[0], str[j]);
        }
    }
    printf("%s\n", y[0]);
    printf("%s\n", z[0]);

    return 0;
}

P1053 回文数

问题描述

解决方法

#include <stdio.h>

int func(int n)
{
    int ori = n;
    int rev = 0;

    while (n > 0)
    {
        int digit = n % 10;
        rev = rev * 10 + digit;
        n /= 10;
    }

    return ori == rev;
}

int main()
{
    int num;
    while (scanf("%d", &num) != EOF)
    {
        if (func(num))
        {
            printf("yes\n");
        }
        else
        {
            printf("no\n");
        }
    }

    return 0;
}

P1036 矩阵转置

问题描述

解决方法

#include <stdio.h>

int main()
{
    int n;
    scanf("%d", &n);
    int a[n][n], b[n][n];
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            scanf("%d", &a[i][j]);
        }
    }
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            b[i][j] = a[j][i];
        }
    }
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            printf("%d ", b[i][j]);
        }
        printf("\n");
    }
    return 0;
}

P1113 选票统计

问题描述

解决方法

#include <stdio.h>

int main()
{
    int input;
    int s[1000];
    s[0] = 0;
    for (int i = 1; i <= 1000; i++)
    {
        s[i] = 0;
    }

    while (scanf("%d", &input) != EOF)
    {
        if (input==-1)
        {
            break;
        }
        
        s[input] += 1;
    }

    for (int i = 1; i <= 1000; i++)
    {
        if (s[i] > 0)
        {
            printf("%d %d\n", i, s[i]);
        }
    }
    return 0;
}

P1070 数据移位

问题描述

解决方法

#include <stdio.h>

int main()
{
    int n, k;

        scanf("%d", &n);
        scanf("%d", &k);
        int a[n];
        for (int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
        }

        for (int i = 0; i < n; i++)
        {
            printf("%d\n", a[k]);
            k++;
            k %= n;
        }
    

    return 0;
}

怎么会有人第一遍看错题 第二遍死循环的啊/哭

P1072 数据合并

问题描述

解决方法

#include <stdio.h>

int main()
{
    int m, n;
    scanf("%d%d", &m, &n);
    int a[m], b[n];
    for (int i = 0; i < m; i++)
        scanf("%d", &a[i]);
    for (int i = 0; i < n; i++)
        scanf("%d", &b[i]);
    int c[m + n], t = 0, i = 0, j = 0;

    while (i < m && j < n)
    {
        if (a[i] < b[j])
            c[t++] = a[i++];
        else if (a[i] > b[j])
            c[t++] = b[j++];
        else 
        {
            c[t++] = a[i++];
            j++;
        }
    }

    while (i < m) c[t++] = a[i++];
    while (j < n) c[t++] = b[j++];

    for (int s = 0; s < t; s++)
    {
        printf("%d", c[s]);
        if (s < t - 1) 
            printf(" ");
    }
    printf("\n"); 

    return 0;
}

P1620 相反数

问题描述

解决方法

#include <stdio.h>

int main()
{
    int N;
    scanf("%d", &N);

    int a[N];
    for (int i = 0; i < N; i++)
    {
        scanf("%d", &a[i]);
    }

    int b[N], t = 0;
    for (int i = 0; i < N; i++)
    {
        b[i] = 0;
    }
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < N; j++)
        {
            if (i != j && a[i] == -a[j] && !b[j])
            {
                t++;
                b[j] = 1;
                break;
            }
        }
    }
    t /= 2;
    printf("%d\n", t);

    return 0;
}

另解（不用变长数组）

#include <stdio.h>
#define MAX_SIZE 100

int main()
{
    int N;
    scanf("%d", &N);

    int arr[MAX_SIZE];
    for (int i = 0; i < N; i++)
    {
        scanf("%d", &arr[i]);
    }
    int count = 0;
    int found[MAX_SIZE] = {0};
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < N; j++)
        {
            if (i != j && arr[i] == -arr[j] && !found[j])
            {
                count++;
                found[j] = 1;
                break;
            }
        }
    }
    count /= 2;
    printf("%d\n", count);
    return 0;
}

P2024 胖头鱼的成绩单

问题描述

#include <stdio.h>

void bubble_sort(int arr[], int n)
{
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n - i - 1; j++)
        {
            if (arr[j] > arr[j + 1])
            {
                int temp = arr[j];
                arr[j] = arr[j + 1];
                arr[j + 1] = temp;
            }
        }
    }
}

int main()
{
    int n;
    scanf("%d", &n);
    int a[n];
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &a[i]);
    }

    bubble_sort(a, n);

    for (int i = n - 1; i >= 0; i--)
    {
        if (a[i] < 70)
        {
            break;
        }
        printf("%d\n", a[i]);
    }

    return 0;
}

P1044 奇偶数排序

问题描述

解决方法

#include <stdio.h>

void bubbleSort(int arr[], int n, int ascending)
{
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = 0; j < n - i - 1; j++)
        {
            if ((ascending && arr[j] > arr[j + 1]) || (!ascending && arr[j] < arr[j + 1]))
            {
                int temp = arr[j];
                arr[j] = arr[j + 1];
                arr[j + 1] = temp;
            }
        }
    }
}

int main()
{
    int sing[10], doub[10];
    int singCount = 0, doubCount = 0;

    for (int i = 0; i < 10; i++)
    {
        int num;
        scanf("%d", &num);
        
        if (num % 2 == 0)
        {
            doub[doubCount++] = num;
        }
        else
        {
            sing[singCount++] = num;
        }
    }

    bubbleSort(sing, singCount, 0);
    bubbleSort(doub, doubCount, 1);

    for (int i = 0; i < singCount; i++)
    {
        printf("%d ", sing[i]);
    }
    for (int i = 0; i < doubCount; i++)
    {
        printf("%d ", doub[i]);
    }
    printf("\n");

    return 0;
}

P1045 奇数求和

问题描述

解决方法

#include <stdio.h>

int main() {
    int nums[10],sum=0;
    for(int i=0;i<10;i++){
        scanf("%d",&nums[i]);
        if (nums[i]%2!=0)
        {
            sum+=nums[i];
        }
    }
    printf("%d\n",sum);
    return 0;
}

P1046 寻找C位数

问题描述

解决方法

#include <stdio.h>

int main() {
    int n;
    scanf("%d", &n);
    int nums[n];
    for (int i = 0; i < n; i++) {
        scanf("%d", &nums[i]);
    }
    printf("%d\n",nums[(n-1)/2]);
    return 0;
}

P1118 ISBN号

问题描述

解决方法

#include <stdio.h>
#include <string.h>

int main()
{
    char a[14], mod[12] = "0123456789X";
    int n;
    scanf("%d", &n);
    getchar();

    while (n--)
    {
        fgets(a, sizeof(a), stdin);
        a[strcspn(a, "\n")] = '\0';

        int i, j = 1, t = 0;
        for (i = 0; i < 12; i++)
        {
            if (a[i] == '-')
            {
                continue;
            }
            t += (a[i] - '0') * j++;
        }
        if (mod[t % 11] == a[12])
        {
            printf("Right\n");
        }
        else
        {
            a[12] = mod[t % 11];
            puts(a);
        }
        getchar();
    }

    return 0;
}

P1248 拐角矩阵

问题描述

解决方法

#include <stdio.h>

int main()
{
    int n;
    scanf("%d", &n);
    int a[n][n];
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            for (int k = 0; k < n; k++)
                if (j >= i && k >= i)
                    a[j][k] = i + 1;

    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
            printf("%d ", a[i][j]);
        printf("\n");
    }
    printf("\n");
    return 0;
}

P1531 字符串排序

问题描述

解决方法

#include <stdio.h>
#include <string.h>

void sortText(char text[][100], int n)
{
    for (int i = 0; i < n - 1; i++)
    {
        for (int j = 0; j < n - i - 1; j++)
        {
            if (strcmp(text[j], text[j + 1]) > 0)
            {
                char temp[100];
                strcpy(temp, text[j]);
                strcpy(text[j], text[j + 1]);
                strcpy(text[j + 1], temp);
            }
        }
    }
}

void getText(char text[][100], int n)
{
    for (int i = 0; i < n; i++)
    {
        fgets(text[i], 100, stdin);
        text[i][strcspn(text[i], "\n")] = '\0';
    }
}

int main()
{
    char text[10][100];
    int n;
    scanf("%d", &n);
    getchar();
    getText(text, n);
    sortText(text, n);
    for (int i = 0; i < n; i++)
    {
        printf("%s\n", text[i]);
    }
    return 0;
}

P1533 字符串中*号的处理

问题描述

解决方法

#include <stdio.h>
#include <string.h>

int main()
{
    char str1[10000], str2[10000];
    int lock = 0, j = 0;
    fgets(str1, 10000, stdin);
    for (int i = 0; i < strlen(str1); i++)
    {
        if (str1[i] == '*' && lock == 1)
        {
            continue;
        }
        if (str1[i] != '*' && lock == 0)
        {
            lock = 1;
        }
        str2[j++] = str1[i];
    }
    str2[j] = '\0';
    printf("%s", str2);
    return 0;
}

P1943 内存不够了

问题描述

解决方法

#include <stdio.h>
#include <stdlib.h>

enum{
    N=100000
};

int main() {
    int n, m;

    scanf("%d %d", &n, &m);
    char **text = malloc(n * sizeof(char *));
    
    for (int i = 0; i < n; i++) {
        text[i] = malloc((N + 1) * sizeof(char));
    }

    int *len = malloc(n * sizeof(int));
    int *find = malloc(m * sizeof(int));

    for (int i = 0; i < n; i++) {
        scanf("%d %s", &len[i], text[i]);
    }
    for (int i = 0; i < m; i++) {
        scanf("%d", &find[i]);
    }
    for (int i = 0; i < m; i++) {
        printf("%s\n", text[find[i] - 1]);
    }

    for (int i = 0; i < n; i++) {
        free(text[i]);
    }
    free(text);
    free(len);
    free(find);

    return 0;
}

P1944 喜提Time Limit Exceeded

问题描述

解决方法

#include <stdio.h>
#include <stdlib.h>

enum {
    N = 20000
};

void swap(int order[], int a1, int a2) {
    int temp = order[a1 - 1];
    order[a1 - 1] = order[a2 - 1];
    order[a2 - 1] = temp;
}

int main() {
    int n, m;
    scanf("%d %d", &n, &m);

    char **text = malloc(n * sizeof(char *));
    for (int i = 0; i < n; i++) {
        text[i] = malloc((N + 1) * sizeof(char));
    }

    int *order = malloc(n * sizeof(int));
    int (*s)[2] = malloc(m * sizeof(*s));

    for (int i = 0; i < n; i++) {
        order[i] = i;
        scanf("%s", text[i]);
    }
    for (int i = 0; i < m; i++) {
        scanf("%d %d", &s[i][0], &s[i][1]);
        swap(order, s[i][0], s[i][1]);
    }
    for (int i = 0; i < n; i++) {
        printf("%s\n", text[order[i]]);
    }

    for (int i = 0; i < n; i++) {
        free(text[i]);
    }
    free(text);
    free(order);
    free(s);

    return 0;
}

有些人，没有TLE，倒是MLE了...

另解 不用动态数组也不会爆内存

#include <stdio.h>

char x[100][20001];
int main()
{
    int index[100];
    int n, t, temp;
    int j, k;
    scanf("%d%d", &n, &t);
    for (int i = 0; i < n; i++)
    {
        scanf("%s", x[i]);
        index[i] = i;
    }
    for (int i = 0; i < t; i++)
    {
        scanf("%d%d", &j, &k);
        temp = index[j - 1];
        index[j - 1] = index[k - 1];
        index[k - 1] = temp;
    }
    for (int i = 0; i < n; i++)
    {
        printf("%s\n", x[index[i]]);
    }

    return 0;
}

P2096 猜数字

问题描述

解决方法

#include <stdio.h>
#include <math.h>

double avg(int arr[], int n)
{
    double sum = 0;
    for (int i = 0; i < n; i++)
    {
        sum += arr[i];
    }
    return sum / n / 2.0;
}

int main()
{
    char name[10][10];
    int guess[10] = {0};
    double average;
    int n;

    scanf("%d", &n);

    for (int i = 0; i < n; i++)
    {
        scanf("%s %d", name[i], &guess[i]);
    }
    
    average = avg(guess, n);

    double min = 100; 
    int record = -1;
    
    for (int i = 0; i < n; i++)
    {
        if (fabs(guess[i] - average) < min)
        {
            min = fabs(guess[i] - average);
            record = i;
        }
    }
    printf("%d %s\n", (int)average, name[record]);
    return 0;
}

忍不住吐槽这题，为什么要跟平均数的一半比啊/恼

第二期完结！！

看完了点个赞再走嘛