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题目描述：
Maxsum
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
思想
用summax变量存储从开始到当前的最大连续和，用k这个变量表示求累加和的开始位置，从第一个数开始遍历，求累加和，当此时累加和大于summax时，就要更新s_id和end_id，并更新summax，如果当前累加和<0了，则需要重新开始算，把当前位置赋值给k，并更新sum为0
代码：
#include<stdio.h>
#include<iostream>
using namespace std;
int arr[100010];
int main(){
int t,cnt=0;
cin>>t;
while(t--){
int n,s_id=0,e_id=0,sum=0,summax=-10000,k=1;
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d",&arr[i]);
}
for(int i=1;i<=n;i++){
sum+=arr[i];
if(sum>summax){
summax=sum;
s_id=k;
e_id=i;
}
if(sum<0){
sum=0;
k=i+1;
}
}
printf("Case %d:\n%d %d %d\n",++cnt,summax,s_id,e_id);
if(t!=0) printf("\n");
}
}
第二种思想
时间复杂度为O（n）的方法：
对于每一个数，将这个数与前几个数的和和下一个数进行比较，如果前者大的话，就取前面的那个，否则就从下一个数开始。这样保证每次加的数的和都是最大的（如果不是最大的，就从下一个数开始重新计算了），所以，特别神奇吧。
代码：
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <stdlib.h>
#include <vector>
#include <algorithm>
#include <math.h>
using namespace std;
const int MAXN=50050;
int n,ans=-2147483647,dp[MAXN],a[MAXN]；
int main()
{
scanf("%d",&n);
for (int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
dp[i]=max(dp[i-1]+a[i],a[i]); //关键
ans=max(ans,dp[i]);
}
printf("%d\n",ans);
return 0;
}