7-1 计算最大值出现的次数

1.题目要求

2.代码实现

#include<stdio.h>
int main()
{
    int arr[1001] = { 0 };
    int n = 0;
    scanf("%d", &n);
    int count = 0;
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &arr[i]);
    }
    for (int i = 0; i < n - 1; i++)//先排序
    {
        int j = 0;
        for (j = 0; j < n - i - 1; j++)
        {
            if (arr[j] > arr[j + 1])
            {
                int tmp = arr[j];
                arr[j] = arr[j + 1];
                arr[j + 1] = tmp;
            }
        }
    }
    for (int i = 0; i < n; i++)//然后计数
    {
        if (arr[i] == arr[n - 1])
           count++;
    }
    printf("%d %d", arr[n - 1], count);
    return 0;
}

7-2 求一批整数中出现最多的个位数字

1.题目要求

2.代码实现

#include<stdio.h>
int main() {
    int num, b, sum = 0, max = 0;
    int flag[10];
    int a[10] = { 0 };
    for (int i = 0; i < 10; i++) {
        flag[i] = -1;
    }
    scanf("%d", &num);
    for (int i = 0; i < num; i++) {
        scanf("%d", &b);
        if (b == 0) {
            a[0]++;
        }
        else {
            while (b > 0) {
                a[b % 10]++;
                b /= 10;
            }
        }
    }
    for (int i = 0; i < 10; i++) {
        if (a[i] > max) {
            max = a[i];
        }
    }
    for (int i = 0; i < 10; i++) {
        if (a[i] == max) {
            flag[sum++] = i;
        }
    }
    printf("%d: ", max);
    for (int i = 0; i < sum; i++) {
        if (flag[i] != -1) {
            printf("%d", flag[i]);
            if (i != sum - 1) {
                printf(" ");
            }
        }
    }
    return 0;
}

7-3 装箱问题

1.题目要求

2.代码实现

#include<stdio.h>
int main()
{
    int m, j, i;
    scanf("%d", &m);
    int a[m];
    int c[10007] = { 0 };//箱内物品总大小
    int b[m];//记录箱子号
    for (i = 0; i < m; i++)
    {
        scanf("%d", &a[i]);
    }
    int k = 0;
    for (i = 0; i < m; i++)
    {
        int p = 0;//设置装箱开关
        for (j = 0; j <= k; j++)
        {
            if ((c[j] + a[i]) <= 100)
            {
                c[j] += a[i];
                b[i] = j + 1;
                p = 1;
                break;
            }
        }
        if (p == 0)//未装进则再开一个箱子
        {
            k++;
            c[k] += a[i];
            b[i] = k + 1;
        }
    }
    for (i = 0; i < m; i++)
    {
        printf("%d %d\n", a[i], b[i]);
    }
    printf("%d", k + 1);
    return 0;
}

7-4 数组-值钱的微信号

1.题目要求

2.代码实现

#include<stdio.h>
int main()
{
    int k = 0;
    int n = 0;
    scanf("%d %d", &k, &n);
    int arr[51] = { 0 };
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &arr[i]);
    }
    for (int i = 0; i < n - 1; i++)//先进行冒泡排序，把消耗的金币按小到大排序
    {
        int j = 0;
        for (j = 0; j < n - i - 1; j++)
        {
            if (arr[j] > arr[j + 1])
            {
                int tmp = arr[j];
                arr[j] = arr[j + 1];
                arr[j + 1] = tmp;
            }
        }
    }
    int sum = 0;
    int count = 0;
    for (int i = 0; i < n; i++)//消耗的越少就要的微信个数越多
    {
        sum += arr[i];
        if (sum < k)
            count++;

    }
    printf("%d", count);
    return 0;
}

7-5 数组-吹泡泡

1.题目要求

2.代码实现

#include<stdio.h>
#define pai 3.14
int main()
{
    int n = 0;
    scanf("%d", &n);
    double flag[1000] = { 0.0 };//记录泡泡的编号的数组
    double arr[1000] = { 0.0 };//收集体积的数组
    for (int i = 0; i < n; i++)
    {
        double x = 0;
        scanf("%lf", &x);
        arr[i] = 1.33 * pai * pow(x, 3);
        flag[i] = 1.33 * pai * pow(x, 3);
    }
    for (int i = 0; i < n - 1; i++)//冒泡排序，按照大到小排序泡泡体积
    {
        int j = 0;
        for (j = 0; j < n - i - 1; j++)
        {
            if (arr[j] < arr[j + 1])
            {
                double tmp = arr[j];
                arr[j] = arr[j + 1];
                arr[j + 1] = tmp;
            }
        }
    }
    for (int i = 0; i < n; i++)
    {
        printf("%.2lf ", arr[i]);//输出体积
        for (int j = 0; j < n; j++)
        {
            if (flag[j] == arr[i])
            {
                  printf("%d\n", j + 1);//输出编号
                flag[j]=0.0;//置零是因为有相同的体积，编号会重复，所以置零
                break;
                
}
              
            
        }
        
    }
    return 0;
}

7-6 数组-数学鬼才

1.题目要求

2.代码实现

#include<stdio.h>
int main()
{
    int n = 0;
    scanf("%d", &n);
    long long sum=1;
    for (int i = 0; i < n; i++)
    {
        sum=1;
        long long x = 0;
        scanf("%lld", &x);
        if(x<2018)
        {
            for(long long j=2;j<=x;j++)
        {
            sum=(sum*j)%2018;
        }
           printf("%lld\n",sum);
    
    }
        else//大于2018的%2018都会是0，会超时，所以直接输出0即可
            printf("0\n");
            
}
        
    return 0;
}