实验6
task1.c
#include <stdio.h> #define N 5 int binarySearch(int* x, int n, int item); int main() { int a[N] = { 2, 7, 19, 45, 66 }; int i, index, key; printf("数组a中的数据:\n"); for (i = 0; i < N; i++) printf("%d ", a[i]); printf("\n"); printf("输入待查找的数据项: "); scanf("%d", &key); index = binarySearch(a, 5, key); if (index >= 0) printf("%d在数组中,下标为%d\n", key, index); else printf("%d不在数组中\n", key); return 0; } int binarySearch(int* x, int n, int item) { int low, high, mid; low = 0; high = n - 1; while (low <= high) { mid = (low + high) / 2; if (item == *(x + mid)) return mid; else if (item < *(x + mid)) high = high - 1; else low = low+ 1; }return -1; }

task2.c
#include <string.h> #include <stdio.h> #include <stdlib.h> void fun(char* a) { int i=0; char* p = a; while (*p && *p == '*') { a[i] = *p; i++; p++; break; } while (*p) { if (*p != '*') { a[i] = *p; i++; }p++; } a[i] = '\0'; } int main() { char s[81]; printf("Enter a string :\n"); gets(s); fun(&s); printf("The string after deleted:\n"); puts(s); return 0; }

task3.c
#include <stdio.h> #include <stdlib.h> #include <string.h> void fun(char* a) { int i = 0; char* t = a, * f = a; char* q = a; while (*t) t++; t--; while (*t == '*') t--; while (*f == '*') f++; while (q < f) { a[i] = *q; q++; i++; } while (q < t) { if (*q != '*') { a[i] = *q; i++; } q++; } while (*q) { a[i] = *q; i++; q++; } a[i] = '\0'; } int main() { char s[81]; printf("Entre a string:\n"); gets(s); fun(s); printf("The sting after deleted:\n"); puts(s); return 0; }

task4.c
#include <stdio.h> #include <string.h> #define N 80 int isPalindrome(char *s); int main() { char str[N]; int flag; printf("Enter a string:\n"); gets(str); flag = isPalindrome(str); if (flag) printf("YES\n"); else printf("No\n"); return 0; } int isPalindrome(char* s) { int i,j; for (i = 0;; ++i) if (*(s+i) == '\0') break; for (j = 0; 2 * j < i; j++) if (*(s + j) != *(s + i - j)) { break; return 0; } return 1; }

task5.c
#include <stdio.h> #define N 80 int count(char* str, char* substr); int main() { char str[N], substr[N]; int n; gets(str); // 输入母串 gets(substr); // 输入子串 n = count(str, substr); // 函数调用 printf("%d\n", n); return 0; } int count(char* str, char* substr) { int i, j, k; int num = 0; for (i = 0; i<N; ++i) for (j=i, k = 0; substr[k] == str[j]; k++, j++) if (substr[k+1] == '\0') { num++; break; } return(num); }

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