问题来源：刘汝佳《算法竞赛入门经典--训练指南》 P61 问题8：
问题描述：给出一个长度为n的序列A1,A2,...,An，求一个连续子序列Ai,Ai+1,...,Aj，使得元素总和最大。
分析：设dp[i]为以i结尾的最大连续和，则d[i] = Max{0,d[i-1]}+Ai;
例题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1003
例题： hdu 1003
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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n个数，求一个连续子序列的最大和(若有多个，取最左边的一个)，并求出最大和的开始节点和终节点。
思路：大致还是上面的思路，只不过多了开始节点和终结点，处理下就可以了
代码：
1 #include "stdio.h" 2 #include "string.h" 3 #define N 100005 4 #define INF 0x3fffffff 5 6 int a[N]; 7 struct node //用结构体存，比较方便 8 { 9 int l,r; 10 int k; 11 }dp[N]; 12 13 int main() 14 { 15 int T,Case; 16 int n; 17 int i,id; 18 scanf("%d",&T); 19 for(Case=1; Case<=T; Case++) 20 { 21 scanf("%d",&n); 22 for(i=1; i<=n; i++) 23 scanf("%d",&a[i]); 24 for(i=0; i<=n; i++) 25 dp[i].l = dp[i].r = dp[i].k = 0; 26 id = 1; 27 dp[1].l = dp[1].r = 1; 28 dp[1].k = a[1]; 29 for(i=2; i<=n; i++) 30 { 31 if(dp[i-1].k >= 0) 32 { 33 dp[i].k = dp[i-1].k+a[i]; 34 dp[i].l = dp[i-1].l; 35 dp[i].r = i; 36 } 37 else 38 { 39 dp[i].k = a[i]; 40 dp[i].l = i; 41 dp[i].r = i; 42 } 43 if(dp[i].k > dp[id].k) 44 id = i; 45 } 46 printf("Case %d: %d %d %d ",Case,dp[id].k,dp[id].l,dp[id].r); 47 if(Case!=T) printf(" "); //格式处理 48 } 49 return 0; 50 }