PAT(Advanced)甲级---1007Maximum Subsequence Sum(25 分)

1007 Maximum Subsequence Sum （25 分）

Given a sequence of K integers { N​1​​, N​2​​, ..., 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:

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

Sample Output:

10 1 4

 

代码:在线处理（扫到一个数就处理一个数）

#include
using namespace std;
const int maxn = 100005;

int main(){
    int K;
    scanf("%d", &K);
    int num[10005];
    for (int i = 0; i < K; i++)scanf("%d", &num[i]);

    int maxsum = -1;
    int first=0, last=K-1;
    for (int i = 0; i < K; i++){ //作为左指针
        int sum = 0;
        int j;
        for (j = i; j < K; j++){
            if (sum+num[j] < 0)break;
            sum += num[j];
            if (sum > maxsum){
                maxsum = sum;
                first = i;
                last = j ;
            }
        }
        i = j;
    }
    if(maxsum<0)maxsum=0;
    printf("%d %d %d", maxsum, num[first], num[last]);

}

 

 

柳神代码（风格不同）：

#include 
#include 
using namespace std;
int main() {
    int n;
    scanf("%d", &n);
    vector v(n);
    int leftindex = 0, rightindex = n - 1, sum = -1, temp = 0, tempindex = 0;
    for (int i = 0; i < n; i++) {
        scanf("%d", &v[i]);
        temp = temp + v[i];
        if (temp < 0) {
            temp = 0;
            tempindex = i + 1;
        } else if (temp > sum) {
            sum = temp;
            leftindex = tempindex;
            rightindex = i;
        }
    }
    if (sum < 0) sum = 0;
    printf("%d %d %d", sum, v[leftindex], v[rightindex]);
    return 0;
}

总结:

• 【动态规划】【最大子列和】

注意

• 测试点4：全负数，maxsum置0。
• 测试点5：例如 -1 0 -3，若maxsum不小于0则会报错。