hdu 1003 Max Sum

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

简单的动态规划，思路不好想。。
但是如果知道大体思路的话就很好理解了。。
代码：
#include
#include
int main(){
	int i,n,m,j,k,t;
	int num[100010];
	int max;
	int sum,temp,temp1,temp2,Case=1;
	scanf("%d",&n);
	while(n--){
		memset(num,0,sizeof(num));
			max=-9999;
			sum=0;
			temp=1;
			temp1=temp2=0;
			scanf("%d",&m);
		for(i=1;i<=m;i++){
			scanf("%d",&num[i]);
			sum+=num[i];
			if(sum>max)
			   {
			   	  max=sum;   //寻找最大值 
			   	  temp1=temp;
			   	  temp2=i;
			   }
			if(sum<0){       //当结果为0时，归0，从下一个数字开始 
				sum=0;
				temp=i+1;
			}
		}
		printf("Case %d:\n",Case++);
		printf("%d %d %d\n",max,temp1,temp2);
		if(n!=0)
		   printf("\n");
	}
	return 0;
}