Max Sum（DP）

Max Sum
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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Status
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

（最后一组数据不用空行）：说白了就是求最大连续子序列的和：

状态方程：dp[i]=max(dp[i-1]+a[i],a[i])表示 以第i个数结尾的最大序列；；； 在这里要解释一下：当dp[i-1]+a[i]>a[i]时

a[i]作为当前序列的结尾加在末尾，否则a[i]作为一个新的列的开端（起点）继续往下搜；

代码：

#include<cstdio>
#include<algorithm>
using namespace std;
int dp[100010];
int a[100010];
int st[100010];
int main()
{
	int t,n;
	scanf("%d",&t);
	int cont=0;		//打脸 
	while(t--)
	{
		scanf("%d",&n);
		for(int i=1;i<=n;i++){
			scanf("%d",&a[i]);
		}
		dp[1]=a[1],st[1]=1;
		for(int i=2;i<=n;i++)
		{
			if(dp[i-1]+a[i]>=a[i]){	//
				dp[i]=dp[i-1]+a[i];
				st[i]=st[i-1];
			}
			else {
				dp[i]=a[i];
				st[i]=i;
			}

		}
		int max=dp[1],sta=1,end=1;
		for(int i=2;i<=n;i++)
		{
			if(dp[i]>max){
				max=dp[i];
				sta=st[i];
				end=i;
			}
			else
			continue;
		}
		printf("Case %d:\n",++cont);
		printf("%d %d %d\n",max,sta,end);
		if(t!=0)
		printf("\n");
	}
	return 0;
}