杭电ACM1003

学习动态规划，就选了一些动态规划的ACM题目练练，这是一道比较简单的题目

我是通过在输入元素的时候就开始计算比较，结构体记录最大子段和的首尾位置

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.

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).

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
  
  

#include<stdio.h>

struct location {
	int first;
	int end;
};

int main(void) {
	int MaxSum , sum;
	int casenum , n;
	int a[100005] , i , j;
	location maxloc , nowloc;
	scanf("%d" , &casenum);
	for(i=1 ; i<=casenum ; i++) {
		scanf("%d" , &n);
		MaxSum = -10001;
		sum = -10001;
		nowloc.first = 1;
		nowloc.end = 1;

		for(j=1;j<=n;j++) {
			scanf("%d" , &a[j]);
			if(sum < 0) {
				sum = a[j];
				nowloc.first = j;
				nowloc.end = j;
				if(sum > MaxSum) {
					MaxSum = sum;
					maxloc.first = nowloc.first;
					maxloc.end = nowloc.end;					
				}
			} else {
				sum += a[j];
				nowloc.end = j;
				if(sum > MaxSum) {
					MaxSum = sum;
					maxloc.first = nowloc.first;
					maxloc.end = nowloc.end;
				}
			}
		}
		printf("Case %d:\n" , i);
		printf("%d %d %d\n" , MaxSum , maxloc.first , maxloc.end);
		if(i != casenum) {
			printf("\n");
		}
	}
 	return 0;
}