hdu 1087

Super Jumping! Jumping! Jumping!

Time
Limit: 2000/1000 MS
(Java/Others)    Memory
Limit: 65536/32768 K (Java/Others)Total Submission(s):
9706    Accepted
Submission(s): 3993
Problem DescriptionNowadays, a kind of chess game called “Super Jumping! Jumping!
Jumping!” is very popular in HDU. Maybe you are a good boy, and
know little about this game, so I introduce it to you now.The game can be played by two or more than two players. It consists
of a chessboard（棋盘）and some chessmen（棋子）, and all chessmen are
marked by a positive integer or “start” or “end”. The player starts
from start-point and must jumps into end-point finally. In the
course of jumping, the player will visit the chessmen in the path,
but everyone must jumps from one chessman to another absolutely
bigger (you can assume start-point is a minimum and end-point is a
maximum.). And all players cannot go backwards. One jumping can go
from a chessman to next, also can go across many chessmen, and even
you can straightly get to end-point from start-point. Of course you
get zero point in this situation. A player is a winner if and only
if he can get a bigger score according to his jumping solution.
Note that your score comes from the sum of value on the chessmen in
you jumping path.Your task is to output the maximum value according to the given
chessmen list. InputInput contains multiple test cases. Each test case is described in
a line as follow:N value_1 value_2 …value_N It is guarantied that N is not more than 1000 and all value_i are
in the range of 32-int.A test case starting with 0 terminates the input and this test case
is not to be processed. OutputFor each case, print the maximum according to rules, and one line
one case. Sample Input

3 1 3
2 
4 1 2
3 4 
4 3 3
2 1 0 Sample Output

4 10 3 这道理的题意是给一个序列要你求它的最大上升序列（可跳）不用连续如序列：1 4 7 3 5 6 那么开另一个数组来保存同下标的“最大值”，每一个都要向前找合法的最大b值，如这里的5可以找的合法值为3 4 1，找到最大后便加上同下标的a值，如：数组a：1   4   7   3   5   6数组b：1   5  12   4  10  16  à这样找到B中的最大值就OK了！ #include<stdio.h>#include<string.h>int
main(){int
n,a[1005],b[1005];int
max,i,j,maxnum;while(scanf("%d",&n),n){maxnum=0;memset(b,0,sizeof(b));for(i=0;i<n;i++){scanf("%d",&a[i]);}for(i=0;i<n;i++){b[i]=a[i];max=0;for(j=0;j<i;j++){ if(a[i]>a[j]){if(b[j]>max){max=b[j];}}}b[i]+=max;if(b[i]>maxnum){maxnum=b[i];}}printf("%d\n",maxnum);}return
0;}