ios学习记录 day7

1.类型 数组名[第一维][第二维] = 初始值;//第一维 第二维 是常量表达式

2.二维数组定义时,可以不指定第一维的长度,但必须给定第二维. 元素个数不足,自动补齐.//int a[][3] = {1,2,3,4,5};

3.字符串 字符型一维数组 | 字符串数组 字符型二维数组

课上练习:

1).将一个二维数组的行和列交换,存储到另外一个数组中去.

    int a[3][4] = {0};
    int b[4][3] = {0};
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 4; j++) {
            a[i][j] = arc4random() % (20 - 10 + 1) + 10;
            printf("%d ",a[i][j]);
        }
        printf(" ");
    }
    printf(" ");
    for (int i = 0; i < 4; i++) {
        for (int j = 0; j < 3; j++) {
            b[i][j] = a[j][i];
            printf("%d ",b[i][j]);
        }
        printf(" ");
    }
2).一个3行4列的二维数组,找出最大元素,输出所在的行和列.
    int a[3][4] = {0};
    int row = 0,col = 0;
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 4; j++) {
            a[i][j] = arc4random() % (20 - 10 + 1) + 10;
            printf("%d ",a[i][j]);
            if (a[row][col] < a[i][j]) {
                row = i;
                col = j;
            }
        }
        printf(" ");
    }
    printf("max = %d 行%d,列%d ",a[row][col],row,col);
(3)char str[3][20] = {"iphone","Android","winPhone"};
    for (int i = 0; i < 3; i++) {
        int length = 0;
        while (str[i][length] != '') {
            printf("%c",str[i][length]);
            length++;
        }
        printf(" ");
    }
(4)char str[3][20] = {"iphone","Android","winPhone"};
    char str1[20] = "";
    
    strcpy(str1, str[1]);
    strcpy(str[1], str[2]);
    strcpy(str[2], str1);
    
    for (int i = 0; i < 3; i++) {
        printf("%s ",str[i]);
    }
5).输入三个字符串,输出最大的那个.

    char str[3][20] = {0};
    int maxIndex = 0;
    for (int i = 0; i < 3; i++) {
        scanf("%s",str[i]);
    }
    for (int i = 0; i < 3; i++) {
        if (strlen(str[maxIndex]) < strlen(str[i])) {
            maxIndex = i;
        }
    }
    printf(" %s ",str[maxIndex]);
6).通讯录有10个人名,按照姓的拼音缩写排列输出
char str[10][20] = {0};
    char temp[20] = {0};
//  随机生成5-15的名字
    for (int i = 0; i < 10; i++) {
        int rLength = arc4random() % (15 - 5 + 1) + 5;//随机产生名字的长度
        for (int j = 0; j < rLength; j++) {
            int r = arc4random() % 2;//真 假
            str[i][j] = r ? arc4random() % ('z'-'a'+1)+'a' : arc4random() % ('Z'-'A'+1)+'A';//真随机小写 假随机大写字母
        }
        str[i][rLength] = '';//名字结束
    }

//    for (int i = 0; i < 10; i++) { /*名字键盘输入
//        scanf("%s",str[i]);
//    }                              */
    
    for (int i = 0; i < 10 - 1; i++) {
        for (int j = 0; j < 10 - i - 1; j++) {
            if (strcmp(str[j],str[j + 1]) > 0) {
                strcpy(temp, str[j]);
                strcpy(str[j], str[j + 1]);
                strcpy(str[j + 1], temp);
                
            }
        }
    }
    for (int i = 0; i < 10; i++) {
        printf("%s ",str[i]);
    }

 作业:

1.输入6个字符串，并对它们按从小到大的顺序排序后输出。

    char a[6][20] = {0};
    char temp[20] = {0};
    for (int i = 0; i < 6; i++) {
        scanf("%s",a[i]);
    }
    for (int i = 0; i < 6 - 1; i++) {
        for (int j = 0; j < 6 - i - 1; j++) {
            if (strcmp(a[j], a[j + 1]) > 0) {
                strcpy(temp, a[j]);
                strcpy(a[j], a[j + 1]);
                strcpy(a[j + 1], temp);
            }
        }
    }
    printf(" ");
    for (int i = 0; i < 6; i++) {
        printf("%s ",a[i]);
    }

2.找出一个二维数组中的“鞍点”，即该位置上的元素在该行中最大，在该列中最小（也可能没鞍点），打印出有关信息。

     int a[3][4] = {
        5,5,5,5,
        1,9,9,9,
        1,9,9,9};
    int have = -1;
    
    
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 4; j++) {
            printf("%d ",a[i][j]);
        }
        printf(" ");
    }
    
    for (int i = 0; i < 3; i++) {
        int mini = 0,maxj = 0;
        for (int j = 0; j < 4; j++) {
            if (a[i][maxj] <= a[i][j]) {
                maxj = j;
            }
        }
        for (int j = 0; j < 4; j++) {
            if (a[i][maxj] == a[i][j]) {
                for (int k = 0; k < 3; k++) {
                    if (a[mini][j] >= a[k][j]) {
                        mini = k;
                    }
                }
                if (mini == i) {
                    printf("a[%d][%d] = %d 是鞍点 ",mini,j,a[mini][j]);
                    have = 0;
                }
            }
        }
    }
    
    if (have) {
        printf("没有鞍点 ");
    }

1.反转一个数组内的元素（例如一个数组内元素为1，2，3，4，5，反转后为5，4，3，2，1）

    int a[5] = {0};
    int b[5] = {0};
    for (int i = 0; i < 5; i++) {
        scanf("%d",&a[i]);
    }
    printf(" ");
    for (int i = 0; i < 5; i++) {
        b[i] = a[4 - i];
        printf("%d ",b[i]);
    }//(1)

    int a[5] = {1,2,3,4,5};
    for (int i = 0; i < 5 / 2; i++) {
        int temp = a[i];
        a[i] = a[5 - i - 1];
        a[5 - i - 1] = temp;
    }
    for (int i = 0; i < 5; i++) {
        printf("%d ",a[i]);
    }//(2)


2.编写一个程序，输入两个包含5个元素的数组，先将两个数组升序排序，然后将这两个数组合并成一个升序数组。
    int a1[5] = {12,48,71,77,88};
    int a2[5] = {21,32,40,55,73};
    int a[10] = {0};
    
    int i = 0,j = 0,k = 0;
    while (i < 5 && j < 5) {
//        if (a1[i] > a2[j]) {
//            a[k] = a2[j];

//            k++;

//            j++;
//        }else{
//            a[k++] = a1[i++];
//        }
        a[k++] = a1[i] < a2[j] ? a1[i++] : a2[j++];
    }//往a里装完了一个数组
    while (i < 5) {
        a[k++] = a1[i++];
    }
    while (j < 5) {
        a[k++] = a2[j++];
    }//往a里装剩下的
    for (int i = 0; i < 10; i++) {
        printf("%d ",a[i]);
    }
3.给定某年某月某日，将其转换成这一年的第几天并输出。
    int Mday[11] = {31,28,31,30,31,30,31,31,30,31,30};
    int year = 0,month = 0,day = 0;
    int sum = 0;
    scanf("%d,%d,%d",&year,&month,&day);
    for (int i = 0; i < month - 1; i++) {
        sum += Mday[i];
    }
    sum += day;
    if (month > 3) {
        if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0) {
            sum++;
        }
    }
    printf("是第%d天 ",sum);


4.char result[50] = {0};
char str1[] = "Lanou ";
char str2[] = “dlios8”;
char str3[] = " is best!";
把str1, str2, str3合并到result数组中。
“Lanou dlios8 is best!”

    char result[50] = {0};
    char str1[] = "Lanou ";
    char str2[] = "dlios8";
    char str3[] = " is best!";
    strcat(result, str1);
    strcat(result, str2);
    strcat(result, str3);
    printf("%s ",result);

5.模拟n个人参加选举的过程，并输出选举结果：假设候选人有四人，分别用A、B、C、D表示，当选某候选人时，直接输入其编号(编号由计算机随机产生)，
若输入的不是A、B、C、D则视为无效票，选举结束后按得票数从高到低输出候选人编号和所得票数。

////    int i=0,n=0,A=0,B=0,C=0,D=0,abandon=0,ran=0;
////    printf("输入参与投票的人数:");
////    scanf("%d",&n);
////    for(i = 0;i < n;i++)
////    {
////        ran = (arc4random() % (4 - 0 + 1) + 0) + 65;
////        switch(ran)
////        {
////            case 'A':
////                A++;
////                break;
////            case 'B':
////                B++;
////                break;
////            case 'C':
////                C++;
////                break;
////            case 'D':
////                D++;
////                break;
////            default :
////                abandon++;
////        }
////        
////    }
////    printf(" A=%d,B=%d,C=%d,D=%d,abandon=%d ",A,B,C,D,abandon);
////    i = 4;
////    while(i-- != 0)
////    {
////        if(A>=B&&A>=C&&A>=D&&A!=-1)
////        {
////            printf("A=%d,",A);
////            A=-1;
////        }
////        if(B>=A&&B>=C&&B>=D&&B!=-1)
////        {
////            printf("B=%d,",B);
////            B=-1;
////        }
////        if(C>=A&&C>=B&&C>=D&&C!=-1)
////        {
////            printf("C=%d,",C);
////            C=-1;
////        }
////        if(D>=A&&D>=B&&D>=C&&D!=-1)
////        {
////            printf("D=%d,",D);
////            D=-1;
////        }
////    }
////    
////    printf("abandon=%d, ",abandon);

6.编写一个程序，要求用户输入一个美金数量， 然后显示出如何用最少的20美元、10美元、5美元和1美元来付款

 int money = 0;
    scanf("%d",&money);
    int array[4] = {20,10,5,1};
    for (int i = 0; i < 4; i++) {
        printf("%-2d元:%2d张 ",array[i],money / array[i]);
        money %= array[i];
    }