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;
}
完全数计算_牛客题霸_牛客网 (nowcoder.com)
#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;
}
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