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): 161361 Accepted Submission(s): 37794
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
代码：
#include <fstream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib> using namespace std; #define EPS 1e-10
#define ll long long
#define INF 0x7fffffff int main()
{
//freopen("D:\\input.in","r",stdin);
//freopen("D:\\output.out","w",stdout);
int T,n,ans,tn,l,r,al,ar,t;
scanf("%d",&T);
for(int tt=;tt<=T;tt++){
scanf("%d",&n);
ans=tn=-INF;
for(int i=;i<=n;i++){
scanf("%d",&t);
if(tn<){
l=r=i;
tn=t;
}else{
tn+=t;
r=i;
}
if(tn>ans){
al=l;
ar=r;
ans=tn;
}
}
printf("Case %d:\n%d %d %d\n",tt,ans,al,ar);
if(tt!=T) puts("");
}
return ;
}