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

题目链接

这篇博文对这类最大子序列和题目的解释很详细，我看了之后感觉通俗易懂，自己照着这个动态规划的思路写了一遍，但是找起点和终点的位置我是按照自己的想法写的，算法会有点复杂。这道题目每组数据之间有空行输出，最后一组之后没有；ans的初始值一定要赋值为一个比较小的负数，因为题目数据是-1000~1000。

AC代码:

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <iomanip>
#include <cctype>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdlib>
#include <sstream>
#include <set>
#include <map>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
typedef long long ll;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5+5;
const double eps = 1e-5;
const double e = 2.718281828459;

int num[maxn];
int maxsum[maxn];

int find_left(int x) {
    int sum = 0, ans = INF;
    for (int i = x; i >= 1; --i) {
        sum += num[i];
        if (sum == maxsum[x]) {
            if (i < ans) {
                ans = i;
            }
        }
    }
    return ans;
}

bool cmp(int a, int b) {
    return a > b;
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    maxsum[0] = 0;
    int t;
    cin >> t;
    for (int Case = 1; Case <= t; ++Case) {
        int n;
        cin >> n;
        for (int i = 1; i <= n; ++i) {
            cin >> num[i];
            maxsum[i] = max(maxsum[i - 1] + num[i], num[i]);
        }
        int ans = -INF, right = 1, left = 1;
        for (int i = 1; i <= n; ++i) {
            if (maxsum[i] > ans) {
                ans = maxsum[i];
                right = i;
                left = find_left(i);
            }
        }
        cout << "Case " << Case << ":" << endl;
        cout << ans << " " << left << " " << right << endl;
        if (Case != t) {
            cout << endl;
        }
    }
    return 0;
}