课程练习三-1003-problem C
2016-05-19 09:17
316 查看
Problem Description
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.<br><br><center><img src=/data/images/1087-1.jpg></center><br><br>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.<br>Your task is to output the maximum value according to the given chessmen list.<br>
Input
Input contains multiple test cases. Each test case is described in a line as follow:<br>N value_1 value_2 …value_N <br>It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int.<br>A test case starting with 0 terminates the input
and this test case is not to be processed.<br>
Output
For each case, print the maximum according to rules, and one line one case.<br>
Sample Input
3 1 3 2
4 1 2 3 4
4 3 3 2 1
0
Sample Output
4
10
3
题意:给定一个序列,求这个序列的最大递增和。
思路:DP,用dp[i]记录下每一个位置序列的最大递增和,然后求max{dp[i]};
代码:
#include<iostream>
#include<stdlib.h>
#include<fstream>
#include<string.h>
#include<algorithm>
using namespace std;
int N;
int arr[1001];
int dp[1001];
int DP()
{
memset(dp,0,sizeof(dp));
dp[1]=arr[0];
int mmax=0;
for(int i=2;i<=N;i++)
{
mmax=0;
for(int j=1;j<i;j++)
{
if(arr[j-1]<arr[i-1])
{
mmax=max(mmax,dp[j]);
}
}
dp[i]=mmax+arr[i-1];
}
int mmm=dp[1];
for(int i=1;i<=N;i++)
{
mmm=max(mmm,dp[i]);
}
return mmm;
}
int main()
{
freopen("C:\\Users\\liuzhen\\Desktop\\11.txt","r",stdin);
while(cin>>N)
{
if(N==0)
break;
for(int i=0;i<N;i++)
{
cin>>arr[i];
}
cout<<DP()<<endl;
}
freopen("con","r",stdin);
system("pause");
return 0;
}
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.<br><br><center><img src=/data/images/1087-1.jpg></center><br><br>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.<br>Your task is to output the maximum value according to the given chessmen list.<br>
Input
Input contains multiple test cases. Each test case is described in a line as follow:<br>N value_1 value_2 …value_N <br>It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int.<br>A test case starting with 0 terminates the input
and this test case is not to be processed.<br>
Output
For each case, print the maximum according to rules, and one line one case.<br>
Sample Input
3 1 3 2
4 1 2 3 4
4 3 3 2 1
0
Sample Output
4
10
3
题意:给定一个序列,求这个序列的最大递增和。
思路:DP,用dp[i]记录下每一个位置序列的最大递增和,然后求max{dp[i]};
代码:
#include<iostream>
#include<stdlib.h>
#include<fstream>
#include<string.h>
#include<algorithm>
using namespace std;
int N;
int arr[1001];
int dp[1001];
int DP()
{
memset(dp,0,sizeof(dp));
dp[1]=arr[0];
int mmax=0;
for(int i=2;i<=N;i++)
{
mmax=0;
for(int j=1;j<i;j++)
{
if(arr[j-1]<arr[i-1])
{
mmax=max(mmax,dp[j]);
}
}
dp[i]=mmax+arr[i-1];
}
int mmm=dp[1];
for(int i=1;i<=N;i++)
{
mmm=max(mmm,dp[i]);
}
return mmm;
}
int main()
{
freopen("C:\\Users\\liuzhen\\Desktop\\11.txt","r",stdin);
while(cin>>N)
{
if(N==0)
break;
for(int i=0;i<N;i++)
{
cin>>arr[i];
}
cout<<DP()<<endl;
}
freopen("con","r",stdin);
system("pause");
return 0;
}
相关文章推荐
- 简单的四则运算
- 数的奇偶性
- ACMer博客瀑布流分析
- ACM程序设计大赛题目分类
- 计算字符串最后一个单词长度
- ACM网址
- 1272 小希的迷宫
- 1272 小希的迷宫
- hdu 1250 大数相加并用数组储存
- 矩阵的乘法操作
- 蚂蚁爬行问题
- 蚂蚁爬行问题
- 求两个数的最大公约数【ACM基础题】
- 打印出二进制中所有1的位置
- 杭电题目---一只小蜜蜂
- HDOJ 1002 A + B Problem II (Big Numbers Addition)
- 初学ACM - 半数集(Half Set)问题 NOJ 1010 / FOJ 1207
- 初学ACM - 组合数学基础题目PKU 1833
- POJ ACM 1002
- POJ 2635 The Embarrassed Cryptographe