专题三acm1001



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

（1）大意：

给出一个数列a[1],a[2],a[3].a[n],我们的工作是计算这个子数列的最大和.例如,给出数列(6,-1,5,4,-7),那么这个数列的最大和就是6 + (-1) + 5 + 4 = 14.

（2）思路：

动态规划的思想，首先求出局部的最大和并保存起来，然后接下来就用本身和前面保存的最大和相比较，选出最大值，到最后就能得到结果。

（3）感想：

思路是有的，但是有思路缺兵也不一定能做对，因为有好多编译错误啥的，关键还是细节，注意容量数组大小，千万就是细节，动态规划千篇一路。

（4）

#include #include using namespace std; int main() {     int i,ca=1,t,s,e,n,x,now,before,max;     scanf("%d",&t);     while(t--)     {        scanf("%d",&n);        for(i=1;i<=n;i++)        {          scanf("%d",&now);          if(i==1)          {             max=before=now;             x=s=e=1;          }          else {              if(now>now+before)              {                 before=now;                 x=i;              }              else before+=now;               }          if(before>max)            {max=before,s=x,e=i;}        }        printf("Case %d:\n%d %d %d\n",ca++,max,s,e);        if(t)     {      printf("\n");     }     }     return 0; }