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[最新] acm-1003

qq41200424 1月前 46


Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 286615    Accepted Submission(s): 68051


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
 


参考求最大子序列,只要处理一下输入输出以及一些细节即可,竞赛题细节很重要,培养了参赛选手的细心程度。

#include <iostream> //采用动态规划,用数组也是可以的,但容易超时 using namespace std; int main() { int T, N, sum, max, a, i, j, left, z, right; cin >> T; for(i = 0; i < T; i++) { cin >> N; for(left = z = right = sum = 0, max = -1001, j = 0; j < N; j++) { cin >> a; sum += a; if(max < sum) { left = z; right = j; max = sum; } if(sum < 0) { z = j + 1; sum = 0;//当总和小于0时就抛弃前面的值 } } cout << "Case " << i + 1 << ":" << endl; cout <<max << " " << ++left << " " << ++right << endl; if(i < T - 1) cout << endl; } return 0; }


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