题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 218106    Accepted Submission(s): 51491

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  

Author

Ignatius.L  

Recommend

思路：很显然，题目就是求最大子段和，只不过，全是负数时不用处理为0。简单的DP，直接上代码吧！注意：此题是样例之间有空格，最后一个样例没有。

附上AC代码：

#include <bits/stdc++.h> using namespace std; const int maxn = 100005; int dp[maxn], l[maxn], r[maxn]; int n; int main(){ int T, cas=0; scanf("%d", &T); while (T--){ scanf("%d", &n); for (int i=1; i<=n; ++i) scanf("%d", dp+i); l[1] = r[1] = 1; for (int i=2; i<=n; ++i){ if (dp[i-1] < 0) l[i] = i; else{ dp[i] += dp[i-1]; l[i] = l[i-1]; } r[i] = i; } int maxsum=dp[1], maxl=1, maxr=1; for (int i=2; i<=n; ++i) if (dp[i] > maxsum){ maxsum = dp[i]; maxl = l[i]; maxr = r[i]; } printf("Case %d:\n%d %d %d\n", ++cas, maxsum, maxl, maxr); if (T) puts(""); } return 0; }