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): 142781 Accepted Submission(s): 33242
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
题目要求我们找一个连续子序列的和最大，并且记录这个序列第一个元素和最后一个元素在原序列中的位置。 AC代码：
#include
using namespace std; int main() { int i,j,k=0,t,n,a,start,end,max,temp; cin>>t; for(i=1;i<=t;i++) { max=-1001,temp=start=k=0; scanf("%d",&n); for(j=0;j
max) { start=k; end=j; max=temp; } if(temp<0)//再往后加会“拖累”后面的和。 { temp=0; k=j+1; } } printf("Case %d:\n",i); printf("%d %d %d\n",max,start+1,end+1); if(i!=t) cout<