hdu 1003 Max Sum 解题报告

链接：http://acm.hdu.edu.cn/showproblem.php?pid=1003

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
  
    

简单的动态规划，最大子串和；

 

  
    

   
      
   
     /*
   
     ******** hdu 1003 ***********
   
     */
   
     
 
   
     /*
   
     ******** 琴心&剑胆 ***********
   
     */
   
     
 
   
     /*
   
     ******** 2011/5/9 ***********
   
     */
   
     

#include
   
     <
   
     stdio.h
   
     >
   
     

   
     int
   
      main(){
 
   
     int
   
      T;
 scanf( 
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     T );
 
   
     int
   
      M
   
     =
   
     T;
 
   
     while
   
     ( T
   
     --
   
      ){
 
   
     int
   
      n;
 scanf( 
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     n );
 
   
     int
   
      a[n],ts
   
     =
   
     0
   
     ,ti
   
     =
   
     0
   
     ,s
   
     =
   
     0
   
     ,bi
   
     =
   
     0
   
     ,bj
   
     =
   
     n
   
     -
   
     1
   
     ,f
   
     =
   
     0
   
     ,max
   
     =-
   
     32767
   
     ,mi;
 
   
     for
   
     ( 
   
     int
   
      i
   
     =
   
     0
   
     ;i
   
     <
   
     n;
   
     ++
   
     i ){
 scanf( 
   
     "
   
     %d
   
     "
   
     ,
   
     &
   
     a[i] );
 
   
     if
   
     (max
   
     <
   
     a[i]){
 max
   
     =
   
     a[i];
 mi
   
     =
   
     i;
 }
 
   
     if
   
     ( a[i]
   
     <
   
     0
   
      )
 f
   
     ++
   
     ;
 ts
   
     +=
   
     a[i];
 
   
     if
   
     (ts
   
     <
   
     0
   
     ){
 ts
   
     =
   
     0
   
     ;
 ti
   
     =
   
     i
   
     +
   
     1
   
     ;
 }
 
   
     if
   
     (s
   
     <
   
     ts){
 s
   
     =
   
     ts;
 bj
   
     =
   
     i;
 bi
   
     =
   
     ti;
 }
 }
 printf(
   
     "
   
     Case %d:\n
   
     "
   
     ,M
   
     -
   
     T);
 
   
     if
   
     ( f
   
     ==
   
     n )
 printf( 
   
     "
   
     %d %d %d\n
   
     "
   
     ,max,mi
   
     +
   
     1
   
     ,mi
   
     +
   
     1
   
      );
 
   
     else
   
     
 printf( 
   
     "
   
     %d %d %d\n
   
     "
   
     ,s,bi
   
     +
   
     1
   
     ,bj
   
     +
   
     1
   
      );
 
   
     if
   
     ( T )
 printf(
   
     "
   
     \n
   
     "
   
     );
 }
}