hdu 1003 Max Sum

        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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

Case 1:14 1 4Case 2:7 1 6

题意概括：找出所给数列的最大子集和，并输出开始坐标和结束坐标。

解题思路：初始sum，max的值为一个数，从第二个数开始遍历每一个数，依次加上sum，若sum的值小于0，然所找的子集开始位置更新，因为sum小于0无论后面如何加都不会是最大值，每次加过之后与max比较大小，如果大于则证明加的这个数可用，让子集的末尾坐标更新。

代码：

#include<stdio.h>int main(){int T,t,i,j,k,n,a[101000],e=0,f,p,max,sum,o;scanf("%d",&T);f=T;while(T--){e++;scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&a[i]);}max=a[0];sum=a[0];o=1;p=1;t=1;for(i=1;i<n;i++){//printf("%d %d\n",sum,max);if(sum<0){sum=0;t=i+1;//t=o;}sum+=a[i];if(sum>max){max=sum;p=i+1;o=t;}}//printf("f%d\n",f);//if(o>p)//o=p;printf("Case %d:\n",e);printf("%d ",max);printf("%d %d\n",o,p);if(e!=f){printf("\n");}}return 0;}