• HDU 1024 Max Sum Plus Plus


    Max Sum Plus Plus

    Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 6721 Accepted Submission(s): 2248


    Problem Description

    Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem.

    Given a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n).

    Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx is not allowed).

    But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^

    Input

    Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn.
    Process to the end of file.

    Output

    Output the maximal summation described above in one line.

    Sample Input

    1 3 1 2 3 2 6 -1 4 -2 3 -2 3

    Sample Output

    6 8
    Hint
    Huge input, scanf and dynamic programming is recommended.

    Author

    JGShining(极光炫影)
     
     
     用两个数组pre[]和now[]来存储前一个状态的值 , now[]存储当前状态的值;
      
     1 #include<stdio.h>
    2 #include<string.h>
    3 #define MAX 1000050
    4 int pre[MAX] , now[MAX] , num[MAX] ; // pre[] 存储前一个状态的最大值 , now[]存储当前状态的值
    5 int max_sum ;
    6 int max ( int a , int b )
    7 {
    8 return a > b ? a : b ;
    9 }
    10 int main ()
    11 {
    12 int n , m ;
    13 while ( scanf("%d%d" , &m , &n) != EOF )
    14 {
    15 for ( int i = 1 ; i <= n ; ++ i )
    16 {
    17 scanf ( "%d" , num + i ) ;
    18 }
    19 memset( pre , 0 , sizeof (pre) ) ;
    20 memset( now , 0 , sizeof (pre) ) ;
    21 for ( int i = 1 ; i <= m ; ++ i )
    22 {
    23 max_sum = -9999999 ;
    24 for ( int j = i ; j <= n ; ++ j )
    25 {
    26 now[j] = max ( now[j-1] + num[j] , pre[j-1] + num[j] ) ;
    27 pre[j-1] = max_sum ;
    28 if ( max_sum < now[j] )
    29 max_sum = now[j] ;
    30 }
    31 }
    32 printf ( "%d\n" , max_sum ) ;
    33 }
    34 return 0 ;
    35 }

      

      

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  • 原文地址:https://www.cnblogs.com/jbelial/p/2126540.html
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