每日一题之 hdu1003 Max Sum

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：

给你一组序列，让你求最大子段和，最大子段和要求是在序列中连续的一段，并记录起始下标和终止位置的下标。我们用一个tmp变量记录加了当前元素之后的值，并和前面所得到的最大值res比较， 如果 $tmp>res$ 那么就把该元素加起来，否则就以当前元素为起点，继续找最大的值。

#include <cstdio>
#include <iostream>
#include <cstring>

using namespace std;

const int maxn = 1e5+5;

long long A[maxn];

int main()
{
    int t;
    cin >> t;
    int cnt = 0;
    while(t--){
        int n;
        cin >> n;
        ++cnt;
        memset(A,0,sizeof(A));
        for (int i = 0; i < n; ++i) {
            cin >> A[i];
        }
        int res = -1e9;
        int st = 0;
        int ed = 0;
        int tmp = 0;
        int l = 0;
        for (int i = 0;i < n; ++i) {

            tmp += A[i];
            if (tmp > res) {
                res = tmp;
                st = l;
                ed = i;
            }
            if (tmp < 0) {
                tmp = 0;
                l = i+1;
            }

        }
        printf("Case %d:\n", cnt);
        printf("%d %d %d\n", res, st+1, ed+1);

        if(t>0) printf("\n");


    }

    return 0;
}