acm Max Sum (diyidao)

C - Max SumTime Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64uSubmit StatusDescriptionGiven a sequence a[1],a[2],a[3]......a, 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 and1000).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 endposition 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

转一下别人的题解：

f
=max{f[n-1]+data
,data
}另加起点，终点的记录。

下面模拟过程：
1.首先，读取第一个数据，令now和max等于第一个数据，初始化pos1,pos2,x位置
2.然后，读入第二个数据，判断
①. 若是now+temp<temp，表示当前读入的数据比之前存储的加上当前的还大，说明可以在当前另外开始记录，更新now=temp
②. 反之，则表示之前的数据和在增大，更新now=now+temp
3.之后，把now跟max做比较，更新或者不更新max的值，记录起始、末了位置
4.循环2~3步骤，直至读取数据完毕

代码：

#include<stdio.h>int main(){int t,n,now,p1,p2,x,j,max,temp;int k=1;scanf("%d",&t);for(int i=1;i<=t;i++){scanf("%d",&n);scanf("%d",&temp);now=max=temp;p1=p2=x=1;for(j=2;j<=n;j++){scanf("%d",&temp);if(now+temp<temp){now=temp;x=j;}elsenow+=temp;if(max<now){max=now;p1=x;//起点由x固定p2=j;//而终点是还没却定的，随j而变化的}}printf("Case %d:\n%d %d %d\n",k++,max,p1,p2);if(i<t) //这里要注意下，最后一组数据后没有空格\n.printf("\n");}return 0;}

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