C是阵列中连续3个数的最大总和

Question

我正在学习C,我的老师给了我一个任务,但我被困住了.在数组中有10个正数,我需要找到具有最高总和的3个连续数字.我在网上找到了这个解决方案,但是如何找到3个连续数字并求它们呢？我该怎么做呢？我应该在数组中查找最大数字并从那里开始吗？我发现的代码在我到目前为止做得更好.

该数组具有:{1,2,3,2,0,5,1,6,0,1},那么3个数字的最大总和将是{ 1,2,3,2,0,5,1,6,0,1},因为它给出了12

#define N_ELEMENTS 10

int main(int argc, const char * argv[]) {

    int a[N_ELEMENTS] = {1, 2, 3, 2, 0, 5, 1, 6, 0, 1};
    int i = 0;
    while(a[i] < 0 && i<N_ELEMENTS) {
        i++;
    }
    if (a[i] < 0) {
        printf ("DEBUG: array with only negative numbers. Print the smallest negative number as the sum and we are done.\n");
    }
    int sum_p=0, sum_n = 0;
    int largest_sum = 0;
    while (i<N_ELEMENTS) {
        if (a[i] > 0) {
            sum_p += a[i];
        }
        else {
            sum_n += a[i];
        }
        if (sum_p+sum_n > largest_sum) {
            largest_sum = sum_p + sum_n;
        }
        if (sum_p+sum_n <= 0) {
            // find the next positive number
            while(a[i] < 0 && i<N_ELEMENTS) {
                i++;
            }
            if (a[i] < 0 || i == N_ELEMENTS) {
                break;
            }
            sum_p = 0;
            sum_n = 0;
        } else {
            i++;
        }
    }
    printf ("DEBUG: The largest consecutive sum = %d\n", largest_sum);
}

Answer 1

你应该做的是尝试每一个可能的连续三个数字,并从中选择最好的类似的东西

int sum = 0;
int idx = 0
for(int i=0;i<size-2;i++){
    if (A[i]+A[i+1]+A[i+2] > sum){
        sum = A[i] + A[i+1] + A[i+2];
        idx = i;
    }
}

答案是A [idx],A [idx + 1],A [idx + 2]