Max Sum
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Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 168618 Accepted Submission(s): 39345
Problem Description
Given a sequence a[1],a[2],a[3]……a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).
Output
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.
Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
Sample Output
Case 1:
14 1 4
Case 2:
7 1 6
代码
#include<iostream>using namespace std;#define MAXSIZE 100000void getMax(int arr[], int len, int &maxValue, int &start, int &end){ int current = 0; int fromIndex = 1; maxValue = -1001; for (int i = 1; i <= len; ++i) { current += arr[i]; if (current >= maxValue) { maxValue = current; start = fromIndex; end = i; } if (current < 0) { current = 0; fromIndex = i + 1; } }}int main(void){ int T; int arr[MAXSIZE] = {}; cin >> T; for (int i = 1; i <= T; ++i) { int len = 0; cin >> len; for (int j = 1; j <= len; j++) { cin >> arr[j]; } int start; int end; int max; getMax(arr, len, max, start, end); cout << "Case " << i << ":" << endl; cout << max << " " << start << " " << end << endl;; if (i < T) cout << endl; memset(arr, '0', sizeof(arr)); } //system("pause"); return 0;}
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