Hduoj1003！【DP】

/*Max Sum
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 142622    Accepted Submission(s): 33185


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

Author
Ignatius.L
*/

题意：求所给数列的最大连续和。

令初始sum为数列第一位，以大于0 的第一个数为起始， 每加上一个数判断sum与max的大小， 记录max以及当前sum的始末位置。

若sum加到某项为负数时，初始化sum为数列的下一项，初始化初位置，并继续求和，求最大的max

 

#include<stdio.h>
int a[100010];
int main()
{
	int i, j, k = 1, fir, la, n;
	scanf("%d", &n);
	while(n--)
	{
		scanf("%d", &a[0]);
		for(i = 1; i <= a[0]; i++)
		scanf("%d", &a[i]);
		int sum , max, x, y;
		max = a[1];
		sum = a[1];
		fir = x = 1;
		la = y = 1;
		for(i = 2; i <= a[0]; i++)
		{
			if( sum < 0)
			{
				sum = a[i];
				x = i;
				y = i;
			}
			else
			{
				sum += a[i];
				y = i;
			}
			if( sum > max )
			{
				max = sum;
				fir = x;
				la = y;
			}
		}
		printf("Case %d:\n%d %d %d\n", k++, max, fir, la);
		if(n) printf("\n");
	}
	return 0;
}

//以下为常规的超时解法......

/*
#include<stdio.h>
int a[100000];
int main()
{
    int T, i, j, fir, la, num = 1, k;
    scanf("%d", &T);
    while(T--)
    {
		int max = -1;
		scanf("%d", &a[0]);
		for( i = 1; i <= a[0] ; i++)
		scanf("%d", &a[i]);
		for( i = 1; i <= a[0]; i++)//start
		{
			for( j = i; j<= a[0]; j++)//end
			{
				int sum = 0;
				for(k = i; k <= j; k++)//max
					sum += a[k];
				if( sum > max )
					{
						max = sum;
						fir = i;
						la = j;
					}
			}
		}
		printf("Case %d:\n%d %d %d\n", num++, max, fir, la);
		if( T )
		printf("\n");
	}
	return 0;
}
*/