Number of times a number appears in the sorting Array

Count the number of times a number appears in the SORT array. Input: each test case contains two rows: the first row has an integer n, indicating the size of the array. 1 <= n <= 10 ^ 6. The second row has n integers, indicating the array elements. Each element is an int. The third row has an integer m, which indicates that there are m queries next. 1 <= m <= 10 ^ 3. There are m rows below, each row has an integer k, indicating the number of queries. Output: For each test case, the m-row output is available. Each row has an integer representing the number of times this number appears in the array. Sample input: 81 2 3 3 3 3 4 513 sample output: 4 idea: Use the bipartite method to locate a specific number, and then scan before and after to determine the number of numbers ~ ^_^ Code AC: [cpp] // use a 2-way method !! Pai_^ # include <stdio. h> # include <stdlib. h> int main () {long int n, I, cou, mid, low, high; int m, * num, k; while (scanf ("% ld ", & n )! = EOF) {num = (int *) malloc (sizeof (int) * n); for (I = 0; I <n; I ++) {scanf ("% d", & num [I]);} scanf ("% d", & m); while (m --) {scanf ("% d ", & k); low = 0; high = n-1; while (low <= high) {mid = (low + high)/2; if (k = num [mid]) {break;} else if (k> num [mid]) {low = mid + 1 ;} else {high = mid-1 ;}} cou = 0; I = mid; while (I <n) {if (num [I] = k) {cou ++; I ++;} else {break;} I = mid-1; while (I >= 0) {if (num [I] = k) {cou ++; I --;} else {break;} printf ("% d \ n", cou) ;}} return 0 ;}