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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

题意：

给定序列个数n及n个数，求该序列的最大连续子序列的和，要求输出最大连续子序列的和以及子序列的首位位置

分析：

  • 常规求解子序列和问题，但是加了首尾位置
  • sum维护当前首尾指向的区间和，maxi记录最大值
    1、如果sum < 0，舍去即可，因为一定导致后面的值变小，不如直接选后面的值
    2、如果sum > maxi，更新最大值
  • 注意结尾换行

AC代码：

#include <iostream>
#include <vector>
#include <utility>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <queue>
#include <stack>
#include <cstdio>
#include <fstream>
#include <set>
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f;
const int MAXN = 1e5 + 100;

int a[MAXN];
int dp[MAXN];
int main() {
  int T;
  cin >> T;
  for (int z = 1; z <= T; z++) {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    int maxi = -INF;
    int start = 0;
    int end = 0;
    int sum = 0;
    int ts = 0, te = 0;
    while (end < n) {
      sum += a[end];
      if (sum > maxi) {
        maxi = sum;
        ts = start;
        te = end;
      }
      if (sum < 0) {
        sum = 0;
        start = end + 1;
      }
      end++;
    }

    if (z != 1)
      cout << endl;
    cout << "Case " << z << ":" << endl
      << maxi << " " << ts + 1 << " " << te + 1 << endl;
  }
  return 0;
}