点击打开链接 密码：syuct
Max Sum
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 1 Accepted Submission(s) : 1
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Problem Description
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
求最长连续子序列和，并输出起始位置和末位置。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=1e5+10;
int a[N];
int inf=1<<29;
int main()
{
int T,n;
scanf("%d",&T);
int ca=1;
while(T--)
{
scanf("%d",&n);
int sum=0,ma=-inf;
int start=1,end=1,k=1;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
if(sum>ma)
{
ma=sum;
start=k;
end=i;
}
if(sum<0)
{
sum=0;
k=i+1;
}
}
printf("Case %d:\n",ca++);
printf("%d %d %d\n",ma,start,end);
if(T!=0)
printf("\n");
}
return 0;
}
