杭电1003（简单dp）

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

求所给的数列最大的子数列之和，同时找出开始和结尾的地方。

开始用数组做的，题目不难，但是超时，然后百度搜了这种方法。

#include 
#include 
int main(void){
	int i,t;
	scanf_s("%d",&t);
	for (i=1; i<=t; i++){
		int j,n,max=-1001,sum=0,tmp=1,*a;
		int first=0,last=0;
		scanf_s("%d",&n);
		a = (int*)malloc(n*sizeof(int));//初始化  计算我们要求的字节数。
		for (j=0; jmax)
			{
				max = sum; 
				first = tmp; 
				last = j+1;
			}
			if (sum<0){
				sum = 0; 
				tmp = j+2;
			}
		}
		printf("Case %d:\n%d %d %d\n",i,max,first,last);
		if (i!=t) printf("\n");
	free(a);//释放malloc的内存
	a = NULL;
	}
	return 0;
}