阵列变数（1）

14 阵列简介

阵列是一群具有相同的资料型态元素集合的资料型态。在记忆体中，一个阵列会使用一段连续的记忆体空间来存放。

阵列的变数定义：

元素资料型态  阵列变数名称 [元素个数] ；   int var[3];

初始化时最少需要指定一个值，而未指定值的元素会被自动指定为0

int counter[6] = {0};

在初始化时，阵列元素个数值可以省略，会以初始化元素个数取代。

int counter[ ] = {0， 0， 0， 0， 0， 0};

 

14.1 骰子点数出现的次数的统计（未使用阵列）

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

int main() {
    srand(time(0));
    int counter1 = 0, counter2 = 0, counter3 = 0;
    int counter4 = 0, counter5 = 0, counter6 = 0;
    int i;
    for (i = 1; i <= 6000; ++i) {
        int dice = rand() % 6 + 1;
        switch (dice) {
            case 1: counter1++; break;
            case 2: counter2++; break;
            case 3: counter3++; break;
            case 4: counter4++; break;
            case 5: counter5++; break;
            case 6: counter6++; break;

        }
    }
    printf("1: %d
", counter1);
    printf("2: %d
", counter1);
    printf("3: %d
", counter1);
    printf("4: %d
", counter1);
    printf("5: %d
", counter1);
    printf("6: %d
", counter1);
    return 0;

}

未使用阵列

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

int main() {
    srand(time(0));
    int counter[6] = {0}; //初始化
    int i;
    for (i = 1; i <= 6000; ++i) {
        int dice = rand() % 6 + 1;
        switch (dice) {
            case 1: counter[0]++; break;
            case 2: counter[1]++; break;
            case 3: counter[2]++; break;
            case 4: counter[3]++; break;
            case 5: counter[4]++; break;
            case 6: counter[5]++; break;

        }
    }
    printf("1: %d
", counter[0]);
    printf("2: %d
", counter[1]);
    printf("3: %d
", counter[2]);
    printf("4: %d
", counter[3]);
    printf("5: %d
", counter[4]);
    printf("6: %d
", counter[5]);
    return 0;

}

1: 982
2: 1014
3: 992
4: 1032
5: 1039
6: 941

Process returned 0 (0x0)   execution time : 13.322 s
Press any key to continue.

 使用阵列后

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

int main() {
    srand(time(0));
    int counter[6] = {0};
    int i,j;
    for (i = 1;i <= 6000; ++i) {
        int dice = rand() % 6 + 1;
        for (i = 1; j <= 6; ++j) {
            if (dice == j) {
                counter[j-1]++;
            }
        }
    }
    for  (j = 1; j <= 6; ++j) {
        printf("%d: %d
", j, counter[j-1]);
    }
    return 0;
}

 这个练习可能很难输出，但是要理解其中的阵列含义

再一次

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

int main() {
    srand(time(0));
    int counter[6] = {0};
    int i, j;
    for (i = 1; i <= 6000; ++i) {
        int dice = rand() % 6 + 1;
            counter[dice-1]++;
    }
    for (j = 1; j <= 6; ++j) {
        printf("%d: %d
", j, counter[j-1]);
    }
    return 0;
}

1: 1001
2: 993
3: 975
4: 1044
5: 1013
6: 974

Process returned 0 (0x0)   execution time : 1.242 s
Press any key to continue.

//  一开始用简单的方法
#include <stdio.h>
int main() {
    int total = 0, id;
    do {
        scanf("%d", &id);
        switch (id) {
            case 1: total += 90;break;
            case 2: total += 75;break;
            case 3: total += 83;break;
            case 4: total += 89;break;
            case 5: total += 71;break;
        }
    } while (id != 0);
    printf("Total: %d
", total);
    return 0;
}

1
2
3
4
5
1
2
3
4
5
0
Total: 816

Process returned 0 (0x0)   execution time : 11.317 s
Press any key to continue.

// 使用阵列后
#include <stdio.h>
int main() {
    int prices[5] = {90, 75, 83, 71};
    int total = 0, id;
    while(1) { 
        scanf("%d", &id);
        if (id == 0) {
            break;
        }
        total += prices[id-1];
    }
    printf("Total: %d
", total);
    return 0;
}


#include <stdio.h>
int main() {
    int prices[5] = {90, 75, 83, 71};
    int total = 0, id;
    do { 
        scanf("%d", &id);
        if (id != 0) {
        total += prices[id-1];
       }
    } while (id != 0);
    printf("Total: %d
", total);
    return 0;
}

 阵列使其变得更为弹性

 

14.5 两颗骰子点数和出现次数统计和的练习（使用阵列）

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

int main() {
    srand(time(0));
    int counter[11] = {0};
    int i;
    for (i = 1; i <= 10000; i++) {
        int dice1 = rand() % 6 + 1;
        int dice2 = rand() % 6 + 1;
        int sum = dice1 + dice2;
        counter[sum-2]++;

    }
    for (i = 2; i <= 12; i++) {
        printf("%d: %d
", i, counter[i-2]);
    }
    return 0;
}

2: 284
3: 597
4: 802
5: 1135
6: 1381
7: 1710
8: 1370
9: 1039
10: 825
11: 573
12: 284

Process returned 0 (0x0)   execution time : 1.121 s
Press any key to continue.

 15.1 查询数字的练习（使用阵列）

#include <stdio.h>
int main() {
    int i,n[5];
    for (i = 1; i <= 5; i++) {
        printf("%d:",i);
        scanf("%d", &n[i-1]);
    }
    while (1) {
        printf("Q: ");
        scanf("%d", &i);
        if (i == 0) {
            break;
        }
        printf("%d
", n[i-1]);

        }
        return 0;
}

1:1
2:23
3:67
4:77
5:89
Q: 5
89
Q: 0

Process returned 0 (0x0)   execution time : 38.987 s
Press any key to continue.

 15.2 查询范围内数字的练习（使用阵列）

#include <stdio.h>

int main() {
    //存数字
    int i,n[10];
    for (i = 1; i <= 10; i++) {
        scanf("%d", &n[i-1]);
    }
    while(1) {
        int l, r;
        printf("L R: ");
        scanf("%d%d", &l, &r);
        if(l == 0 && r == 0) {
            break;
        }
        printf("Ans:  ");
        for (i = 0; i < 10; i++ ) {
            if (n[i] >= l && n[i] <= r) {
                printf("%d", n[i]);
            }
        }
        printf("
");
    }
    return 0;

}

07 17 27 37 47 57 67 77 87 97
L R: 70 80
Ans:  77
L R: 1 90
Ans:  71727374757677787
L R: 77 97
Ans:  778797
L R: 0 0

Process returned 0 (0x0)   execution time : 39.275 s
Press any key to continue.

15.3 查詢上限內最大數字的練習 (使用陣列)

#include <stdio.h>

int main()
{
    int i, n[10], q;
    for (i = 1; i <= 10; i++)
    {
        scanf("%d", &n[i-1]);
    }
    // 用数组来存10个数字，就是这样存的。
    while (1) {// 无限回圈
        int max_i = -1;    // 每次它都要归零
        printf("Q: ");
        scanf("%d", &q);
        if (q == 0) {
            break; // 终止条件
        }
        for (i = 0; i < 10; i++) {
            if (n[i] <= q && (max_i == -1 || n[i] > n[max_i])) {// 这里的||一个成立就够了
                max_i = i; // 索引号
            }
        }
        if (max_i != -1) {
            printf("%d
", n[max_i]);
        }
    }
    return 0;
}

12 23 34 56 67 78 98 87 97 99
Q: 10
Q: 15
12
Q: 100
99
Q: 13
12
Q: 11
Q: 0

Process returned 0 (0x0)   execution time : 25.608 s
Press any key to continue.

15.4 查詢最接近數字的練習 (使用陣列)

#include <stdio.h>

int main()
{
    int i,q,n[10];
    for (i = 1; i <= 10; i++) {
        scanf("%d", &n[i-1]);
    }
    while (1) {
    printf("Q: ");
    scanf("%d", &q);
    if (q == 0) {
        break;
    }
    int nearest_n = n[0], nearest_d = abs(q-n[0]);
    for (i = 1; i < 10; i++) {
        int d =abs(q-n[i]);
        if (
            d < nearest_d || (d == nearest_d && n[i] < nearest_n)) {
                nearest_d = d;
                nearest_n = n[i];

            }

    }
    printf("%d
", nearest_n);

   }
   return 0;
}

13 23 78 77 37 30 40 50 83 99
Q: 25
23
Q: 34
37
Q: 9
13
Q: 0

Process returned 0 (0x0)   execution time : 30.611 s
Press any key to continue.

15.5  写数字统计长条图的练习（使用阵列）

#include <stdio.h>

int main()
{

    int i, j, n, b[10] = {0};
    for (i = 1; i <= 10; i++) {
        scanf("%d", &n);
        b[(n-1)/10]++; //存数
    }
    for (i = 1; i <= 10; i++) {
        printf("%3d: ", i*10);
        for (j = 1; j <= b[i-1]; j++) { //这里的b[i-1]表示一个区间里有几个星星
            printf("*");
        }
        printf("
");

    }return 0;

}

12 24 35 45 55 65 75 88 98 99
 10:
 20: *
 30: *
 40: *
 50: *
 60: *
 70: *
 80: *
 90: *
100: **

Process returned 0 (0x0)   execution time : 36.454 s
Press any key to continue.

 15.6 求小范围内的众数 （使用阵列）

当次数一样多的时候，数大的输出。

#include <stdio.h>

int main()
{
    int i, n, b[10] = {0};
    for (i = 1; i <= 10; i++) {
        scanf("%d", &n);
        b[n]++;

    }
    int ans = 0; //ans表示出现最多次数的那个数
    for (n = 1; n < 10; n++) {
        if (b[n] >= b[ans]) {
            ans = n;

        }
    }
    printf("Ans: %d
", ans);
    return 0;
}

1 3 4 4 5 4 4 6 7 8
Ans: 4

Process returned 0 (0x0)   execution time : 18.670 s
Press any key to continue.