Candy and candy ppt

Candy and candy ppt

Description There are n Children waiting in queue to ask you for candy. Every child has a cute value. You must give each child at least one candy, and the cute and high-value child must have more candy than the adjacent child. Note: If the cute values of two adjacent persons are the same, there is no limit. Ask how much candy you have to prepare at least. For example, there are five children whose cute values are 3, 40, 2, 1, and 10. Then the minimum solution should be 1, 3, 2, 1, 2, 9 in total About Input The first line is a positive integer, n, indicating the number of children. (N <= 100000) There are n positive integers in the second row. The I integer represents the cute value of the I child (within the int range ). Numbers are separated by a space. About output A positive integer with the minimum number of sweets to be prepared. Example Input 53 40 2 1 10 Example output 9

#include <stdio.h>#include <stdlib.h>int main(){int n = 0, j;int *p = 0, *q = 0;int t = 0;scanf("%d", &n);p = (int *)calloc(n, sizeof(int));q = (int *)calloc(n, sizeof(int));for (j = 0; j < n; j++){scanf("%d", &p[j]);}q[0] = 1;for (j = 1; j < n; j++){if (p[j] <= p[j - 1])q[j] = 1;elseq[j] = q[j - 1] + 1;}for (j = n - 2; j >= 0; j--){if (p[j] > p[j + 1] && q[j] <= q[j + 1])q[j] = q[j + 1] + 1;}for (j = 0; j < n; ++j){t += q[j];}printf("\n%d\n", t);return 0;}