zoukankan      html  css  js  c++  java
  • Super Jumping! Jumping! Jumping!——E

                                                                                   

                                 E. Super Jumping! Jumping! Jumping!

    Time Limit: 1000ms

    Memory Limit: 32768KB

    64-bit integer IO format:      Java class name:

    Nowadays, 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.
     

    Input

    Input 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.
     

    Output

    For 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


    题意:求最大递增子段和,
    #include <cstdio>
    #include <iostream>
    #include<cstring>
    using namespace std;
    const int inf = 999999999;
    int main()
    {
        int n,a[1005],sum[1005],maxn;
       while(scanf("%d",&n)&&n)
       {
           memset(sum,0,sizeof(sum));
           for(int i = 1;i<=n;i++)
            scanf("%d",&a[i]);
           for(int i=1;i<=n;i++)
           {
               maxn=-inf;
               for(int j=0;j<i;j++)
                 if(a[i]>a[j])
                   maxn=max(maxn,sum[j]);
               sum[i]=maxn+a[i];
               }
               maxn=-inf;
               for(int i=1;i<=n;i++)
               {
                   if(sum[i]>maxn) maxn=sum[i];
               }
               printf("%d
    ",maxn);
       }
        return 0;
    }
  • 相关阅读:
    Android自定义属性,format详解
    设置EditText是否可编辑
    Android中Application设置全局变量以及传值
    android:ellipsize属性的含义
    Java集合框架List,Map,Set等全面介绍
    二十款漂亮的CSS字体样式,让你受用非浅
    IE浏览器不见了
    display属性值:
    用CSS设置多个背景、背景渐变、指定背景大小
    [转]IE6双倍边距——IE布局BUG集锦
  • 原文地址:https://www.cnblogs.com/fenhong/p/5288984.html