01-复杂度2 Maximum Subsequence Sum

Given a sequence of K integers { N1, N2, …, N**K }. A continuous subsequence is defined to be { N**i, N**i+1, …, N**j } where 1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (≤10000). The second line contains K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:

1 2	10 -10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4

分析

求最大子列各，及子列的开始、结束位置

编码：

#include <stdio.h>
#define MAXSIZE 10000

int maxSub(int A[], int N, int *index);

int main_p5() {
    int number, i, total, sum, index, arr[MAXSIZE];
    scanf("%d", &number);
    for (i = 0; i < number; i++) {
        scanf("%d", &arr[i]);
    }
    total = maxSub(arr, number, &index);
    // 如果全为负数
    if (total < 0){
        total = 0;
        index = number -1;
        i = 0;
        printf("%d ", total);
        printf("%d %d\n", arr[i], arr[index]);
        return 0;
    }
    printf("%d ", total);
    sum = 0;
    // 从最大子列和最末位index向前遍历，至sum等于total，得到开始索引i
    for (i = index; i >= 0; i--) {
        sum += arr[i];
        if (sum == total) {
            break;
        }
    }
    printf("%d %d\n", arr[i], arr[index]);

    return 0;
}

// 在线处理，即时处理算法
int maxSub(int A[], int N, int *index) {
    int ThisSum = 0, MaxSum = -1;
    int i;
    for (i = 0; i < N; i++) {
        ThisSum += A[i]; // 向右累加
        if (ThisSum > MaxSum) {
            MaxSum = ThisSum; // 发现更大和则更新当前结果
            *index = i;
        } else if (ThisSum < 0) { // 如果当前子列和为负
            ThisSum = 0; // 则不可能使后面的部分和增大，抛弃之
        }
    }
    return MaxSum;
}

小结：

根据在线算法求得最大子列和，并用index标记子列的最末位置index
根据index向前遍历数组，并累加子列和，至与最大和相等，记录索引i
输入最大子列和，子列开始位置i，子列结束位置index