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
1 #include<stdio.h>
2 #include<string.h>
3 #include<algorithm>
4 using namespace std;
5 int main(){
6 int t,n,a[100010],cnt=1,dp[100010];
7 scanf("%d",&t);
8 int q=t;
9 while(t--){
10 scanf("%d",&n);
11 memset(dp,0,sizeof(dp));
12 for(int i=1;i<=n;i++){
13 scanf("%d",&a[i]);
14 dp[i]=a[i];
15 }
16 int maxx=dp[1];
17 int t=1;
18 for(int i=1;i<=n;i++){
19 dp[i]=max(dp[i],dp[i-1]+a[i]);
20 if(dp[i]>maxx){
21 maxx=dp[i];
22 t=i; //记录最长子序列的末位置
23 }
24 }
25 int sum=0,tt;
26 for(int i=t;i>=1;i--){
27 sum+=a[i];
28 if(sum==maxx){ //找最长子序列的初始位置
29 tt=i;
30 }
31 }
32 printf("Case %d:\n",cnt++);
33 printf("%d %d %d\n",maxx,tt,t);
34 if(q!=(cnt-1))
35 printf("\n");
36 }
37 }
原文地址：https://www.cnblogs.com/cake-lover-77/p/10197474.html