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. ^_^
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.
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 }