Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 142281 Accepted Submission(s): 33104
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
问题的想法是《编程之美》上的，修改了一下提交就ok了。。
思路：
考虑数组的第一个元素a[0],以及最大的一段数组（a[i].....a[j]）跟a[0]之间的关系，有一下几种情况：
1. 当0=i=j时，元素a[0]本身构成和最大的一段。
2.当0=i<j时，和最大的一段以a[0]开始。
3.当0<i时，元素a[0]跟和最大的一段没有关系。
假设已经知道（a[1]....a[n-1]）中的最大的一段数组之和为All[1]，并且已经知道了（a[1]......a[n-1]）中包含a[1]的和最大的一段数组为start[1]。
那么有以上情况可以得出（a[0].....a[n-1]）中问题的解All[0]是三种情况的最大值max(a[0],a[0]+start[1],All[1]).
so问题符合无后效性，用动态规划方法可解决。
1 #include<iostream> 2 using namespace std; 3 int main() 4 { 5 int t,len,m=0,h,n,w,l,r,p,e,i; 6 cin>>t;h=t; 7 while(t--) 8 { 9 m++; 10 l=r=p=e=0; 11 cin>>len; 12 int a[len]; 13 for(i=0;i<len;i++) 14 cin>>a[i]; 15 n=a[0],w=a[0];l=0;p=0; 16 for(i=1;i<len;i++) 17 { 18 if(a[i]>n+a[i]){ 19 n=a[i]; 20 l=i; 21 r=i; 22 } 23 else { 24 n=n+a[i]; 25 r=i; 26 } 27 if(n>w){ 28 w=n; 29 e=r; 30 p=l; 31 } 32 } 33 cout<<"Case "<<m<<":"<<endl<<w<<" "<<p+1<<" "<<e+1<<endl; 34 if(m!=h)cout<<endl; 35 } 36 return 0; 37 }