C语言程序设计PPT电子课件教案-05_用数组存储信息.ppt

lesson 5 用数组存储信息,5,本课内容,二维数组及应用,整型数组初始化,数组元素逆置,求数组极值、平均值,数组元素平移,一维数组8-1,2006.9.35若要求定义具有10个int型元素的一维数组a，则以下定义语句中错误的是 a)#define n 10 b)#define n 5 int an; int a2*n; c)int a5+5; d)int n = 10, an;,2010.3.25下列能正确定义一维整型数组的语句是 a)int num02008; b)int num.1; c)int n = 2008; d)#define n 2008 int numn; int numn;,一维数组8-2,2010.9.27下列定义数组的语句中，正确的是 a)int n = 10; b)#define n 10 int xn; int xn; c)int x010; d)int x ;,2010.9.28要定义有5个元素的数组，错误的是 a)int i = 5, ai; b)int b5 = 0; c)int c = 0,0,0,0,0; d)int d2+3;,2008.9.21以下选项对m数组，引用错误的是 int m = 5,4,3,2,1, i = 4; a)mm0 b)m-i c)m2*2 d)mmi,一维数组8-3,2007.9.142010.9.92011.3.132011.9.9以下程序的运行结果是 int i, a5 = 0; for (i = 1; i = 4; i+) ai = ai-1 * 2 + 1; printf(“%d“, ai); ,13715,一维数组8-4,2005.4.23以下程序运行后，j的值是 int p8 = 11,12,13,14,15,16,17,18; int i = 0, j = 0; while (i+ 7) if (pi % 2) j += pi;,45,2007.4.14以下程序的运行结果是 int x6 = 1,2,3,4,5, a5, i; for (i = 0; i 5; i+) ai = xi + xi+1; printf(“%d“, ai); ,35795,一维数组8-5,2010.3.29以下程序运行后s的值是 int a5 = 1,2,3,4,5, i, s = 0; int b5 = 0,2,1,3; for (i = 0; i 5; i+) s = s + abi;,11,2008.4.31以下程序的运行结果是 int s12 = 1,2,3,4,4,3,2,1,1,1,2,3; int c5 = 0, i; for (i = 0; i 12; i+) csi+; for (i = 1; i 5; i+) printf(“%d“, ci);,4332,一维数组8-6,2010.9.132011.9.13以下程序运行结果是 int a2, i, j; for (i = 0; i 2; i+) ai = 0; for (i = 0; i 2; i+) for (j = 0; j 2; j+) aj = ai + 1; printf(“%d“, a1);,3,一维数组8-7,2009.9.29程序运行后a数组的内容是 int a = 2,3,4,5, i; for (i = 0; i 4; i+) switch (i % 2) case 0: switch (ai % 2) case 0: ai+; break; case 1: ai-; break; case 1: ai = 0; ,3050,2005.4.30以下叙述中错误的是 a)对于double类型数组，不可以直接用数组名对数组进行整体输入或输出 b)数组名代表的是数组所占存储区的首地址，其值不可改变 c)当程序执行中，数组元素的下标超出所定义的下标范围时，系统将给出“下标越界“的出错信息 d)可以通过赋初值的方式确定数组元素的个数,一维数组8-8,求数组相邻元素之和,2007.4.14求数组x中各相邻两个元素的和，依次存放到a数组中，并输出。 int x10, a9, i; for (i = 0; i 10; i+) scanf(“%d“, ,i = 1,xi-1,求平均分,填空实现程序的功能：计算a数组中10个学生的平均分，并输出。 int a10, i; _; for (i = 0; i 10; i+) scanf(“%d“, ,double s = 0,%f,输出低于平均分的成绩,填空实现程序的功能：输出数组中低于平均分的学生的成绩。 int a10, i; double s = _; for (i = 0; i 10; i+) scanf(“%d“, ,0,ai/10.0,二维数组,二维数组的定义,int a34 = 1,2,3,4,5,6,7,8,9,10, 11,12; 或 int a34 = 1,2,3,4,5,6,7,8, 9,10,11,12;,输出二维数组,for (i = 0; i 3; i+) for (k = 0; k 4; k+) printf(“%2d,“, aik); printf(“n“); ,1, 2, 3, 4,5, 6, 7, 8,9,10,11,12,输出：,二维数组定义3-1,2006.4.162011.9.26以下数组定义错误的是 a)int x 3 = 0; b)int x23 = 1,2,3,4,5,6; c)int x 3 = 1,2,3,4,5,6; d)int x23 = 1,2,3,4,5,6;,例以下数组定义中正确的是 a)int n = 5, x n = 0; b)int x3 = 1,2,3; c)int x 3 = 1,2,3,4,5,6; d)int x 3 = 1,2,3,4,5,6;,二维数组定义3-2,2008.4.24以下数组定义中错误的是 a)int x 3 = 0,1,1,2,3; b)int x 3 = 1,2,3,4,5,6; c)int x4 = 1,2,3,1,2,3, 1,2,3,1,2,3; d)int x 3 = 1,2,3,4;,2007.9.31若有定义语句：int a36; 按在内存中的存放顺序，a数组的第10个元素是 a)a04 b)a13 c)a03 d)a14,二维数组定义3-3,2008.4.26设有定义语句：int a23; double i = 1; 则以下选项对a数组元素正确引用的是 a)a2!1 b)aii c)a1.22.1 d)a12!1,2004.9.27以下能正确定义二维数组的是 a)int a 3; b)int a 3 = 2*3; c)int a 3 = ; d)int a23 = 1,2,3,4;,二维数组8-1,2006.4.8以下程序运行后，s的值是 int a33 = 1,2,9,3,4,8,5,6,7; int i, s = 0; for (i = 0; i 3; i+) s += aii + ai3-i-1;,30,2010.3.11以下程序的运行结果是 int a33 = 1,2,3,4,5,6,7,8,9; int b3, i; for (i = 0; i 3; i+) printf(“%d“, bi = ai2 + a2i);,101418,二维数组8-2,2011.9.28以下程序的运行结果是 #define n 4 int xnn = 1,2,3,4,5,6,7,8, 9,10,11,12,13,14,15,16; int yn, i; for (i = 0; i n; i+) yi = xii - xin-1-i; for (i = 0; i n; i+) printf(“%d“, yi);,-3-113,二维数组8-3,2010.3.9以下程序的运行结果是 int i, j, a 3 = 1,2,3,4,5,6,7,8,9; for (i = 0; i 3; i+) for (j = i; j 3; j+) printf(“%d“, aij);,123569,2010.3.30以下程序运行后，t的值是 int a33 = 0,1,2,0,1,2,0,1,2, t = 1; for (i = 0; i 3; i+) for (j = 1; j = i; j+) t *= aiaji;,4,二维数组8-4,2008.9.35以下程序的运行结果是 int y4, i, x 4 = 1,2,3,4, 5,6,7,8,9,10; for (i = 0; i 4; i+) yi = xii; printf(“%d“, yi); ,1070,2006.4.34以下程序的运行结果是 int i, t 3 = 9,8,7,6,5,4,3,2,1; for (i = 0; i 3; i+) printf(“%d“, t2-ii);,357,二维数组8-5,2007.4.13设有以下定义语句： int a 3 = 0,1,2; 则数组元素a12的值为,0,2008.9.36以下程序的运行结果是 int a44 = 1,2,3,4,11,12,13,14, 21,22,23,24,31,32,33,34; int i, m = a00; for (i = 1; i m) m = aii; printf(“%d“, m);,34,二维数组8-6,2004.9.10以下程序运行后，s的值是 int a44 = 1,2,3,4,2,3,1,0, 5,7,4,2,3,9,1,7; int i = 0, j = 0, s = 0; while (i+ 4) if (i=2 | i=4) continue; j = 0; do s += aij; j+; while (j4); ,26,二维数组8-7,2008.9.12程序运行结果是 int a44, n = 0, i, j; for (j = 0; j = j; i-) n+; aij = n; for (i = 0; i 4; i+) for (j = 0; j = i; j+) printf(“%3d“, aij); printf(“n“); ,4 3 7 2 6 9 1 5 8 10,二维数组8-8,2007.4.35以下程序的运行结果是 int a44 = 1,4,3,2,8,6,5,7, 3,7,2,5,4,8,6,1, i, k, t; for (i = 0; i 3; i+) for (k = i+i; k 4; k+) if (aii akk) t = aii; aii = akk; akk = t; for (i = 0; i 4; i+) printf(“%d“, a0i);,6432,数组周边元素,求nn的二维数组周边元素的平均值。 #define n 4 int i, k, ann = 1,2,3,4,5,6,7,8,9, 10,11,12,13,14,15,16; double s = 0; for (i = 0; i n; i+) for (k = 0; k n; k+) if (_) s += aik; printf(“%f“, s / (4*n-4);,i=0 | k=0 | i=n-1 | k=n-1,数组元素平移,0,1,2,3,4,t = a0;,0,1,2,3,4,0,a4 = t;,for (i = 1; i = 4; i+) ai-1 = ai;,数组元素平移,填空实现程序的功能：把一维数组中下标从0到m的数组元素平移到数组的最后。 int a5 = 1,3,5,7,9, i, k, t, m; scanf(“%d“, ,0,4,数组元素平移,2011.3.11已知a所指的数组中有n个元素。函数fun的功能是，将下标k(k0)开始的后续元素全部向前移动一个位置。请填空。 void fun(int an, int k) int i; for (i = k; i n; i+) a = ai; ,i - 1,fibonacci数列,a0 = 0, a1 = 1; for (i = 2; i 20; i+) ai = ai-1 + ai-2;,0,1,1,2,3,5,8,13,21,34,0,1,1,2,3,5,8,13,21,34,55,折半查找,折半查找2-1,i,k,m,v = 9;,m = (i + k) / 2;,m = 3,2,4,6,9,12,15,18,21,34,55,67,折半查找2-2,2,4,6,9,12,15,18,21,34,55,67,i,k,m,v = 19;,m = (i + k) / 2;,无此数据!,筛选法求素数,for (i = 2; i 100; i+) ai = i; for (i = 2; i 100; i+) if (ai != 0) printf(“%d,“, i); for (k = 2*i; k 100; k += i) ak = 0; ,2,3,4,5,6,7,8,9,10,11,12,杨辉三角,魔术方阵,17,1,8,15,24,23,7,14,16,5,4,13,20,22,6,10,19,21,3,12,11,25,2,9,18,数据分类统计,统计数组a中的各年龄段的人数。09岁的人数放在b0中，1019岁的人数放在b1中，以此类推，100岁及以上的放在b10中。 int a20, b11, i; for (i = 0; i =100 ? b10+ : b_+;,0,ai/10,数制转换,填空实现程序的功能：将十进制整数m转换成k进制（2 = k = 9）数的数字输出。例如，如果m = 8，k = 2，则应该输出1000。 int a20, i, m, k; scanf(“%d%d“, ,m % k,k,塔形方阵,螺旋方阵,把二维数组 a 各行之和分别放入 b 数组，各列之和分别放入 c 数组。,3,2,4,1,4,2,5,2,3,9,for (i = 0; i 3; i+) for (k = 0; k 4; k+) bi += aik, ck += aik;,求二维数组行列之和,3,1,5,6,7,8,3,5,选择法排序,0,1,2,3,4,6,5,3,8,4,6,5,8,6,3,4,6,5, m = i; for (k = i+1; k 5; k+) if (ak am) m = k; t = am, am = ai, ai = t; ,i = 0;,for (i = 0; i 4; i+),冒泡法排序,for (i = 0; i ak) t = ai, ai = ak, ak = t;,0,1,2,3,4,5,2,3,8,4,5,4,8,5,3,5,5,2,左边界,右边界,右边界+1,冒泡法排序,将a数组中n个整数，除第一个和最后一个整数外，将其余n-2个整数按降序排列。 int an = 6,3,7,4,2,5,9, i, k, t; for (i = _; i n-2; i+) for (k = i+1; k n-1; k+) if (ai _ ak) t = ai, ai = ak, ak = t; for (i = 0; i n; i+) printf(“%d,“, ai);,1,矩阵转置,矩阵转置，若输入10 20 30 则输出10 40 70 (即行列互换) 40 50 60 20 50 80 70 80 90 30 60 90 #define n 3 int i, k, t, ann = 10,20,30,40,50, 60,70,80,90; for (i = 0; i