ECNU839数据结构2005答案

ecnu-05-1-1-1#define N 1000struct DataType{int acctNum;char name[10];float balance;};int main(){int i;struct DataType record = { 0, "", 0.0 };FILE *cfp;if ((cfp = fopen("file.dat", "w")) == NULL)printf("File could not be opened.\n");else{for (i = 0; i < N; i++)fwrite(&record, 1, sizeof(DataType), cfp);fclose(cfp);}}ecnu-05-1-1-2void sign_on(char str[], char *password){int a = 0;while ((strcmp(str, password) != 0) && ++a < 10){printf("Enter password please:");gets(str);}}int main(){char pwd[] = { "7788" };char str[] = { "5566" };sign_on(str, pwd);}ecnu-05-1-2-1int f(int a){static int x = 1;return (x*=a);}int main(){int i;for (i = 1; i < 5; i ++)printf("%d %d\n", i, f(i));return 1;}// output:// 1 1// 2 2// 3 6// 4 24// 没有5 120 !!!ecnu-05-1-2-2int main(){int c;if ((c = getchar()) != '0'){main();printf("%c", c);}return 1;}// input: This is 4096<CR>// output: 4 si sihTecnu-05-1-2-3int main()int i, j;for (i = 0; i < 3; i ++){j = 3;switch(i){case 0:while(j){case 1:printf("i=%d, ", i);case 2:printf("j=%d, ", j); j--;default:;}}printf("---\n");}return 1;}// output:// i=0, j=3, i=0, j=2, i=0, j=1, ---// i=1, j=3, i=1, j=2, i=1, j=1, ---// j=3, i=2, j=2, i=2, j=1, ---//ecnu-05-1-3-1int test (unsigned bits){unsigned i, mask = 1 << 15, count = 0;for (i = 0; i < 16; i++, bits <<=1){if ((bits & mask) == mask){count ++;}}return (count % 2 ? 0:1);}int main(){int result = test(123);// 检测bits 中含1的个数为奇数个返回1，否则返回0}ecnu-05-1-4-1void sort(int *p, int n) // 递归!!! 选一个最小值，和第一个首元素交换{int myMin = 0xFFFF, myMinIndex = 0;for (int i = 0; i < n; i ++){if (p[i] < myMin){myMin = p[i];myMinIndex = i;}}// swapp[myMinIndex] = p[0];p[0] = myMin;if (n > 1){sort(&p[1], n - 1);}}int main(){int a[] = { 1, 9, 7, 5, 3 };sort(a, sizeof(a)/sizeof(int));return 1;}ecnu-05-1-4-2long total(int n, int a) // days, things{int sum;if (n > 1){sum = total(n - 1, (a + 1)*2);}else{sum = a;}return sum;}int _tmain(int argc, _TCHAR* argv[]){/*x0 = nday1: x1 = x0/2-1day2: x2 = x1/2-1dayn: xn = a*/int days = 3, things = 1;printf("total = %d", total(days, things)); return 0;}ecnu-05-1-4-3enum colour{red,yellow,blue,};colour c[3] = { red, yellow, blue };void swap(colour **pc, int index1, int index2) {colour *tmp = (colour *)pc[index1];pc[index1] = pc[index2];pc[index2] = tmp;}colour inspect(colour **pc, int index)return *pc[index];}void holland(colour **pc, int count){int swapped = 0;for (int i = 0; i < count; i ++){swapped = 0;for (int j = count - 1; j > i; j --){if (inspect(pc, j - 1) > inspect(pc, j)){swap(pc, j - 1, j);swapped = 1;}}if (swapped == 0){ // 没有乱序break;}}return ;}int main(){colour *pc[8] = { &c[1], &c[0], &c[0], &c[2], &c[1], &c[2], &c[0], &c[2] };printf("Original: ");for (int i = 0; i < 8; i ++){printf("%d ", *pc[i]);}holland(pc, 8);printf("\r\n");printf("Result : ");for (int i = 0; i < 8; i ++){printf("%d ", *pc[i]);}return 0;}数据结构ecnu-05-填1.(n+1)/2n^2/2(1-成功的概率)不成功: n+(n-1)+(n-2)+...+1/n=(n+1)*n/2=(n+1)/2成功:(n+1)/2注：求查找不成功时的平均查找长度，分母为表长2.(r-f+n)%n(Length=(rear－front+Maxsize)%Maxsize)3.d3.e广义表：head取a1->表头，tail取a2...an->表尾A=((a,b,c),(d,e,f))B=tail(A)=(d,e,f)head(B)=d最后在做任何运算也应该是d吧？4.8(主对角线元素个数)4.8i+j=18j+i => i=0,j=0 & i=17, j=7 => 有2个解5.99,无穷(n0=n2+1,n0=50 => n2=49 => n=n0+n2=99)6.n\2+17.6,8(B树元素最少为3个，最多为5个)8.不为0和∞的元素个数(0代表自己，无穷代表不连通)ecnu-05-二-1带头结点的链表：统一处理头结点和一般结点环形链表：访问到最后一个结点可以直接返回到第一个结点带头结点的环形链表：同上双向链表：可以向前、向后遍历ecnu-05-二-20 A->1->4->^1 B->2->3->^2 C->0->1->^3 D->^4 E->1->^ecnu-05-二-3错。

同义词的线性探测再散列的元素位置是连续的，但在到达最后一个位置后，会从0开始计算，存储位置不连续(散列表分为1.开放地址法2.拉链法)同义词在被删除后仅仅标记为已删除，方便下次搜索，并没有回收空间，因此可能不连续ecnu-05-二-4()/ \() ()/\ /\60()90()/\ /\40()5050/\1030ecnu-05-二-5a f e|\/ /||/\ / |b c decnu-05-三-1(1)稳定算法：void bucketSort(int p[], int n){int bucket[m] = { 0 };for (int i = 0; i < n; i ++){bucket[p[n]] ++; // 放入桶中}for (int i = 0, n = 0; i < m; i ++){for (int j = 0; j < bucket[i]; j ++){p[n++] = i; // 出}}(2)时间：O(n)~O(n^2)空间：O(m) (m个不同的值)(3)类似于基数排序(Radix Sort)，但排序范围受桶个数的限制ecnu-05-三-2// 返回值：0xFF:不是平衡查找树(A VL); -1 or 0 or 1 平衡因子int judgeBin(BiTree *bt){int l = 0, r = 0;if (bt->left != NULL){if (bt->left->data > bt->data) // 左子树的值必须小{return 0xFF;}l = judgeBin(bt->left) + 1; // 计算平衡因子if (l == 0xFF) // 0xFF 标记失败return 0xFF;}if (bt->right != NULL){if (bt->right->data < bt->data){return 0xFF;}r = judgeBin(bt->right) + 1;if (l == 0xFF) // 0xFF 标记失败return 0xFF;}if ((l - r) > 1 || (l - r) < 1)return 0xFF;return l > r ? l : r; // 返回结点高度}。