杭电1003

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Problem DescriptionGiven 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.

InputThe 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).

OutputFor 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 Input2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5 
Sample OutputCase 1:
14 1 4

Case 2:
7 1 6

注意事项：
1：输入数据有可能为负数；
2：输出格式要注意。

#include <stdio.h> 
int a[100010]; 
int main(void)
{
<span style="white-space:pre">	</span>int n,m,k = 1,i,pos,start,end;
<span style="white-space:pre">	</span>scanf("%d",&n);
<span style="white-space:pre">	</span>while(n--)
<span style="white-space:pre">	</span>{
<span style="white-space:pre">		</span>scanf("%d",&m);
<span style="white-space:pre">	</span>for(i = 1; i <= m; i++)
<span style="white-space:pre">	</span>{
<span style="white-space:pre">		</span>scanf("%d",&a[i]);
<span style="white-space:pre">	</span>}
<span style="white-space:pre">	</span>int s = 0,max = a[1];
<span style="white-space:pre">	</span>pos = start = end = 1;
<span style="white-space:pre">	</span>for(i = 1; i <= m; i++)
<span style="white-space:pre">	</span>{
<span style="white-space:pre">		</span>if(s < 0)
<span style="white-space:pre">		</span>{
<span style="white-space:pre">			</span>s = a[i];
<span style="white-space:pre">			</span>pos = i;
<span style="white-space:pre">		</span>}
<span style="white-space:pre">		</span>else
<span style="white-space:pre">			</span>s += a[i];
<span style="white-space:pre">		</span>if(max < s)
<span style="white-space:pre">		</span>{
<span style="white-space:pre">			</span>max = s;
<span style="white-space:pre">			</span>start = pos;
<span style="white-space:pre">			</span>end = i;
<span style="white-space:pre">		</span>}
<span style="white-space:pre">	</span>}
<span style="white-space:pre">	</span>printf("Case %d:\n%d %d %d\n",k++,max,start,end);
<span style="white-space:pre">	</span>if(n > 0)
<span style="white-space:pre">		</span>printf("\n");
<span style="white-space:pre">	</span>}
<span style="white-space:pre">	</span>return 0;
}

杭电1003

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原文地址：http://blog.csdn.net/u013224148/article/details/46973767