Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 306494 Accepted Submission(s): 72850
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
分析：
第一次是用暴力解法，时间复杂度为O(n*n)，果然TLE了！！
上网查了之后才明白这是个动态分配的题。大体思路是依次遍历data数组，若sum>=0，则令sum+=data[i]，否则sum=data[i]，然后比较sum和max，若sum>max，则令max=sum，并修改相应的子列起初下标start和截至下标end。这样一次遍历下来之后就找到了和最大的子列。时间复杂度为O(n)。
注意输出格式，输出每个Case的结果后空一行（除最后一个Case）。
下面给出AC代码。
#include<cstdio>
using namespace std;
int main(){
int T,cas=0;
int data[100005];
scanf("%d",&T); //输入T（测试用例个数）
while(T--){
int max=-1e8,sum=0,start=0,end=0,s=0; //初始时给max一个很小的值
printf("Case %d:\n",++cas);
int n;
scanf("%d",&n); //输入序列大小
for(int i=0;i<n;i++){
scanf("%d",&data[i]);
if(sum>=0){
sum+=data[i];
}
else{
sum=data[i];
s=i; //s为当前序列的起始下标
}
if(sum>max){
max=sum;
start=s;
end=i;
}
}
printf("%d %d %d\n",max,start+1,end+1);
if(T!=0)
printf("\n");
}
return 0;
}