485. 最大连续 1 的个数 - 力扣（LeetCode）

#include <stdio.h>  int findMaxConsecutiveOnes(int* nums, int numsSize) {  if (numsSize == 0) return 0; // 如果数组为空，返回0  int maxCount = 0;  // 最大连续1的个数  int currentCount = 0;  // 当前连续1的个数  for (int i = 0; i < numsSize; i++) {  if (nums[i] == 1) {  // 如果当前元素是1，增加当前连续1的个数  currentCount++;  // 更新最大连续1的个数  if (currentCount > maxCount) {  maxCount = currentCount;  }  } else {  // 如果当前元素是0，重置当前连续1的个数  currentCount = 0;  }  }  return maxCount;  
}  int main() {  int nums[] = {1, 1, 0, 1, 1, 1};  // 示例二进制数组  int numsSize = sizeof(nums) / sizeof(nums[0]);  // 计算数组的大小  int maxConsecutiveOnes = findMaxConsecutiveOnes(nums, numsSize);  printf("The maximum number of consecutive 1s is: %d\n", maxConsecutiveOnes);  return 0;  
}

 

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#include <stdio.h>  // 函数用于计算一个数的所有真因子之和  
int sumOfDivisors(int num) {  int sum = 0;  for (int i = 1; i < num; i++) {  if (num % i == 0) {  sum += i;  }  }  return sum;  
}  int main() {  int n;  printf("Enter the number of test cases: ");  scanf("%d", &n);  for (int t = 0; t < n; t++) {  int num;  printf("Enter a positive integer: ");  scanf("%d", &num);  if (sumOfDivisors(num) == num) {  printf("%d is a perfect number.\n", num);  } else {  printf("%d is not a perfect number.\n", num);  }  }  return 0;  
}