题目链接

http://vjudge.net/contest/132391#problem/G

Description

standard input/output
Statements

— It' s a good game, — Princess said pensively. It was clear that she was thinking about something else.

— They like to play various games here in Castles Valley. And they invent ones themselves. Say, my friend Knight played with a princess a game some time ago, — Dragon thought it was a good idea o tell Princess about another game, if, perhaps, previous game was seemed no interesting for her.

Princess A. offered Knight to play a game of numbers. She puts down the number zero on a sheet of paper. Let us call this number acurrent result.

Further steps of princess A. and Knight are described below. She calls any positive integer and Knight says what she must do with this number: to add it to the current result or subtract it from the current result.

Princess A. performs the action and calculates a new value. This value becomes the new current result.

Princess A. wants that current result to be not less than zero and not greater than k at any time. The game finishes when an action makes the result out of the range or when a sequence of n numbers, which princess A. conceived, exhausts.

Knight managed to learn the sequence of n numbers that princess A. guessed, and now he wants the game to last as long as possible.

Your task is to compute maximum possible number of actions which Knight is able to perform during the game.

Input

The first line contains integers n and k (1 ≤ n ≤ 1000,  1 ≤ k ≤ 1000) — the size of sequence which princess A. conceived and an upper bound for a current result which must not be exceeded.

The second line contains n integers c1, c2, ..., cn (1 ≤ cj ≤ k) — the sequence which princess A. conceived.

Output

In the first line print integer d — maximum possible number of actions, which Knight is able to perform during the game.

Print d symbols "+" and "-" in the second line. Symbol at jth position specifies an action which is applied to jth number in the princess' sequence. If multiple answers exist, choose any of them.

Sample Input

Input
2 5
3 2
Output
2
++
Input
5 5
1 2 3 4 5
Output
4
++-+ 题意:输入n,k 然后输入n个正整数(每个数小于等于k 大于0)从第一个数开始加上或者减去这个数,使得当前的算式值在0~k之间,求这个算式的最大长度,
并输出这个算式的运算符; 思路:DP,定义dp[i][j]表示由前i个数能否得到j,能则dp[i][j]=1,否则为0; 代码如下:
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <map>
#include <vector>
using namespace std;
int a[];
char s[];
bool dp[][]; int main()
{
int n,k;
while(scanf("%d%d",&n,&k)!=EOF)
{
for(int i=;i<=n;i++)
scanf("%d",&a[i]);
memset(dp,,sizeof(dp));
dp[][a[]]=;
int i,j;
for(i=;i<=n;i++)
{
int f=;
for(j=;j<=k;j++)
{
if(dp[i-][j])
{
if(j+a[i]<=k) { dp[i][j+a[i]]=; f=; }
if(j>=a[i]) { dp[i][j-a[i]]=; f=; }
}
}
if(f==) break;
}
printf("%d\n",i-); s[]='+'; s[i]='\0'; i--;
for(j=;j<=k;j++)
if(dp[i][j]) break; for(;i>=;i--)
{
if(j>=a[i]&&dp[i-][j-a[i]]) { s[i]='+'; j=j-a[i];}
else { s[i]='-'; j=j+a[i]; }
}
puts(s+);
}
return ;
}
    

Gym 100703G---Game of numbers(DP)的更多相关文章

  1. URAL 1586 Threeprime Numbers(DP)

    题目链接 题意 : 定义Threeprime为它的任意连续3位上的数字,都构成一个3位的质数. 求对于一个n位数,存在多少个Threeprime数. 思路 : 记录[100, 999]范围内所有素数( ...

  2. POJ1338Ugly Numbers(DP)

    http://poj.org/problem?id=1338 第一反应就是DP,DP[i] = min{2*DP[j], 3*DP[k], 5*DP[p] j,k,p<i};于是枚举一下0-i- ...

  3. Codeforces 403D: Beautiful Pairs of Numbers(DP)

    题意:转换模型之后,就是1~n个数中选k个,放到一个容量为n的背包中,这个背包还特别神奇,相同的物品摆放的位置不同时,算不同的放法(想象背包空间就是一个长度为n的数组,然后容量为1的物体放一个格子,容 ...

  4. 【Gym - 101002F】Mountain Scenes(dp)

    Mountain Scenes Descriptions 给你一个长度为n的丝带,一个宽w一个高h 的 格子,用丝带去填充格子,这填充后只需要满足至少有一列的丝带长度与其他格子不同即可.丝带可以不全部 ...

  5. 【gym102394B】Binary Numbers(DP)

    题意:From https://blog.csdn.net/m0_37809890/article/details/102886956 思路: 可以发现转移就是右上角的一个区间前缀和 std只要开1倍 ...

  6. 【CF55D】Beautiful numbers(动态规划)

    [CF55D]Beautiful numbers(动态规划) 题面 洛谷 CF 题解 数位\(dp\) 如果当前数能够被它所有数位整除,意味着它能够被所有数位的\(lcm\)整除. 所以\(dp\)的 ...

  7. LightOJ 1033 Generating Palindromes(dp)

    LightOJ 1033  Generating Palindromes(dp) 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid= ...

  8. lightOJ 1047 Neighbor House (DP)

    lightOJ 1047   Neighbor House (DP) 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87730# ...

  9. UVA11125 - Arrange Some Marbles(dp)

    UVA11125 - Arrange Some Marbles(dp) option=com_onlinejudge&Itemid=8&category=24&page=sho ...

随机推荐

  1. EF架构~为IEnumerable接口添加增删查等操作,原因是IEnumerable导航属性更放心

    回到目录 对EF开发来说,导航属性肯定都用过,事实上,它是由VS IDE工具根据你的数据库关系结构自动生成的外键属性,在类视图中可以看到相关属性,它是以外键表名来标识的,如果是一对多的关系,那么,它会 ...

  2. Node.js入门:模块机制

    CommonJS规范      早在Netscape诞生不久后,JavaScript就一直在探索本地编程的路,Rhino是其代表产物.无奈那时服务端JavaScript走的路均是参考众多服务器端语言来 ...

  3. React(三)组件的生命周期

    Component Specs and LifeCycle <div id="app"></div> <script src="bower_ ...

  4. OpenCascade BRep Format Description (2)

    OpenCascade BRep Format Description eryar@163.com 摘要Abstract:本文结合OpenCascade的BRep格式描述文档和源程序,对BRep格式进 ...

  5. 深入理解PHP内核(六)哈希表以及PHP的哈希表实现

    原文链接:http://www.orlion.ga/241/ 一.哈希表(HashTable) 大部分动态语言的实现中都使用了哈希表,哈希表是一种通过哈希函数,将特定的键映射到特定值得一种数据 结构, ...

  6. Huge Page 是否是拯救性能的万能良药?

    本文将分析是否Huge Page在任何条件下(特别是NUMA架构下)都能带来性能提升. 本博客已经迁移至: http://cenalulu.github.io/ 为了更好的体验,请通过此链接阅读: h ...

  7. PHP的学习--PHP的闭包

    php的闭包(Closure)也就是匿名函数,是PHP5.3引入的. 闭包的语法很简单,需要注意的关键字就只有use,use是连接闭包和外界变量. $a = function() use($b) {} ...

  8. nodejs操作mongodb

    一.下载地址 https://www.mongodb.com/download-center#community 二.控制台操作mongodb 1.安装完后添加环境变量. 2.在某个根目录下新建dat ...

  9. Android Studio1.4.x JNI开发基础 - 简单实例

    接上一篇,搭建好基于Android Studio的环境之后,编写native代码相对来说也比较简单了.在Android上编写Native代码和在Linux编写C/C++代码还是有区别,Native代码 ...

  10. Sql Server优化之索引提示----我们为什么需要查询提示,Sql Server默认情况下优化策略选择的不足

    环境: Sql Server2012 SP3企业版,Windows Server2008 标准版 问题由来: 最近在做DB优化的时候,发现一个存储过程有非常严重的性能问题, 由于整个SP整体逻辑是一个 ...