PTA | Maximum Subsequence Sum

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   作者: 陈越 单位: 浙江大学 时间限制: 200 ms 内存限制: 64 MB 代码长度限制: 16 KB   1 #include <stdio.h> 2 3 void MaxSubseqSum(int A[], int N) 4 { 5 int ThisSum = 0, MaxSum = 0; 6 int beg, end; 7 int tmp = A[0]; 8 int max = A[0]; 9 for (int i = 0; i < N; i++) 10 { 11 ThisSum += A[i]; 12 if (A[i] > max) 13 { 14 max = A[i]; 15 } 16 if (ThisSum > MaxSum) 17 { 18 MaxSum = ThisSum; 19 beg = tmp; 20 end = A[i]; 21 } 22 else if (ThisSum < 0) 23 { 24 ThisSum = 0; 25 tmp = A[i + 1]; 26 } 27 } 28 if (max < 0) 29 { 30 beg = A[0]; 31 end = A[N - 1]; 32 } 33 else if (max == 0) 34 { 35 beg = end = 0; 36 } 37 printf("%d %d %d", MaxSum, beg, end); 38 } 39 40 int main() 41 { 42 int N; 43 int num[100000]; 44 scanf("%d", &N); 45 for (int i = 0; i < N; ++i) 46 { 47 scanf("%d", &num[i]); 48 } 49 MaxSubseqSum(num, N); 50 51 return 0; 52 }

 

提交结果

测试点提示结果耗时内存0sample换1个数字。有正负，负数开头结尾，有并列最大和答案正确2 ms384KB1最大和序列中有负数答案正确1 ms384KB2并列和对应相同i但是不同j，即尾是0答案正确2 ms256KB31个正数答案正确2 ms384KB4全是负数答案正确2 ms256KB5负数和0答案正确2 ms256KB6最大和前面有一段是0答案正确2 ms256KB7最大N答案正确3 ms384KB  

转载于:https://www.cnblogs.com/lkpp/p/PTAMaximumSubsequenceSum.html