1244. Gentlemen

Time limit: 0.5 second
Memory limit: 64 MB
Let's remember one old joke:
Once a gentleman said to another gentleman:
— What if we play cards?
— You know, I haven't played cards for ten years…
— And I haven't played for fifteen years…
So, little by little, they decided to resurrect their youth. The first gentleman asked a servant to bring a pack of cards, and before starting playing out weighed in his hand the pack.
— It seems to me, one card is missing from the pack… — he said and gave the pack to the other gentleman.
— Yes, the nine of spades, — the man agreed.
An incomplete pack of cards is given. The program should determine which cards are missing.

Input

The first line contains a positive integer, which is the weight in milligrams of the given incomplete pack. The second line contains an integer N, 2 ≤ N ≤ 100 — the number of cards in the complete pack. In the next N lines there are integers from 1 to 1000, which are the weights of the cards in milligrams. It's guaranteed that the total weight of all cards in the complete pack is strictly greater than the weight of the incomplete pack.

Output

If there is no solution, then output the single number 0. If there are more than one solutions, then you should write −1. Finally, if it is possible to determine unambiguously which cards are missing in the incomplete pack as compared to the complete one, then output the numbers of the missing cards separated with a space in ascending order.

Samples

input output
270
4
100
110
170
200
2 4
270
4
100
110
160
170
-1
270
4
100
120
160
180
0
Problem Author: Alexander Petrov
Problem Source: Ural State University Personal Programming Contest, March 1, 2003
Difficulty: 284
 
题意:给出x 。然后n个ai,问哪几个ai能够组成x。若无解,输出0,不止一个解,输出-1,否则输出那几个数的编号
分析:背包。 路径。
 #include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <ctime>
#include <iomanip>
using namespace std;
typedef long long LL;
typedef double DB;
#define For(i, s, t) for(int i = (s); i <= (t); i++)
#define Ford(i, s, t) for(int i = (s); i >= (t); i--)
#define Rep(i, t) for(int i = (0); i < (t); i++)
#define Repn(i, t) for(int i = ((t)-1); i >= (0); i--)
#define rep(i, x, t) for(int i = (x); i < (t); i++)
#define MIT (2147483647)
#define INF (1000000001)
#define MLL (1000000000000000001LL)
#define sz(x) ((int) (x).size())
#define clr(x, y) memset(x, y, sizeof(x))
#define puf push_front
#define pub push_back
#define pof pop_front
#define pob pop_back
#define ft first
#define sd second
#define mk make_pair
inline void SetIO(string Name)
{
string Input = Name+".in",
Output = Name+".out";
freopen(Input.c_str(), "r", stdin),
freopen(Output.c_str(), "w", stdout);
} inline int Getint()
{
int Ret = ;
char Ch = ' ';
bool Flag = ;
while(!(Ch >= '' && Ch <= ''))
{
if(Ch == '-') Flag ^= ;
Ch = getchar();
}
while(Ch >= '' && Ch <= '')
{
Ret = Ret * + Ch - '';
Ch = getchar();
}
return Flag ? -Ret : Ret;
} const int N = , M = ;
int n, m, Arr[N];
int Dp[M], G[M];
bool Visit[N];
vector<int> Ans; inline void Input()
{
scanf("%d", &m);
scanf("%d", &n);
For(i, , n) scanf("%d", Arr + i);
} inline void Search(int x)
{
if(!x) return;
Visit[G[x]] = ;
Search(x - Arr[G[x]]);
} inline void Solve()
{
Dp[] = ;
int Max = ;
For(i, , n)
{
Max += Arr[i];
if(Max > m) Max = m;
Ford(j, Max, Arr[i])
if(Dp[j - Arr[i]])
{
Dp[j] += Dp[j - Arr[i]];
if(!G[j]) G[j] = i;
}
} if(Dp[m] > ) puts("-1");
else if(!Dp[m]) puts("");
else
{
Search(m);
For(i, , n)
if(!Visit[i]) Ans.pub(i);
int Length = sz(Ans);
Rep(i, Length - ) printf("%d ", Ans[i]);
printf("%d\n", Ans.back());
}
} int main()
{
#ifndef ONLINE_JUDGE
SetIO("C");
#endif
Input();
Solve();
return ;
}
 

ural 1244. Gentlemen的更多相关文章

  1. DP URAL 1244 Gentlemen

    题目传送门 /* 题意:已知丢失若干卡片后剩余的总体积,并知道原来所有卡片的各自的体积,问丢失的卡片的id DP递推:首先从丢失的卡片的总体积考虑,dp[i] 代表体积为i的方案数,从dp[0] = ...

  2. 递推DP URAL 1244 Gentlemen

    题目传送门 /* 题意:给出少了若干卡片后的总和,和原来所有卡片,问少了哪几张 DP:转化为少了的总和是否能有若干张卡片相加得到,dp[j+a[i]] += dp[j]; 记录一次路径,当第一次更新的 ...

  3. URAL 1244. Gentlemen(DP)

    题目链接 这题不难啊...标记一下就行了.表示啥想法也没有. #include <cstring> #include <cstdio> #include <string& ...

  4. URAL 1244. Gentlemen (DP)

    题目链接 题意 : 给出一幅不完全的纸牌.算出哪些牌丢失了. 思路 : 算是背包一个吧.if f[j]>0  f[j+a[i]] += f[j];然后在记录一下路径. #include < ...

  5. URAL 1244

    题目大意:给出一个正整数M,给出N个正整数ai,让你在这些数中挑出一些数组成M的一个划分,如果符合条件的划分数超过两个,输出:-1,如果没有输出:0,如果有且仅有一个:则按顺序输出剩下的数的序号. 例 ...

  6. URAL DP第一发

    列表: URAL 1225 Flags URAL 1009 K-based Numbers URAL 1119 Metro URAL 1146 Maximum Sum URAL 1203 Scient ...

  7. 【51Nod 1244】莫比乌斯函数之和

    http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1244 模板题... 杜教筛和基于质因子分解的筛法都写了一下模板. 杜教筛 ...

  8. 51nod 1244 莫比乌斯函数之和

    题目链接:51nod 1244 莫比乌斯函数之和 题解参考syh学长的博客:http://www.cnblogs.com/AOQNRMGYXLMV/p/4932537.html %%% 关于这一类求积 ...

  9. 后缀数组 POJ 3974 Palindrome && URAL 1297 Palindrome

    题目链接 题意:求给定的字符串的最长回文子串 分析:做法是构造一个新的字符串是原字符串+反转后的原字符串(这样方便求两边回文的后缀的最长前缀),即newS = S + '$' + revS,枚举回文串 ...

随机推荐

  1. gem install factory_girl

    文章是从个人博客转过来的,  可以直接访问 iwangzheng.com https://github.com/thoughtbot/factory_girl https://github.com/t ...

  2. 三款SDR平台对比:HackRF,bladeRF和USRP

    这篇文章是Taylor Killian今年8月发表在自己的博客上的.他对比了三款平价的SDR平台,认为这三款产品将是未来一年中最受欢迎的SDR平台.我觉得这篇文章很有参考价值,简单翻译一份转过来.原文 ...

  3. poj3026(bfs+prim)

    The Borg is an immensely powerful race of enhanced humanoids from the delta quadrant of the galaxy. ...

  4. ubuntu12 apache2部署多个网站的方法

    由于对apache2的配置不是很精通.吃了不少瘪. 这里总结下,希望对碰到同样问题的人带来帮助: 假设需求是有两个站点,site1和site2.其端口分别为80和8080. 1. 配置监听端口 修改 ...

  5. vim实用技巧

    <1> 删除空格: :% s/ //gi    #正则为一个空格,替换为空,全局匹配. <2> 删除空行: :g /^\n*$/ d  #g为global  正则为:行开始+换 ...

  6. Counting Bits

    Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the ...

  7. 12.从上往下遍历二元树[LevelOrderOfBinaryTree]

    [题目] 输入一颗二元树,从上往下按层打印树的每个结点,同一层中按照从左往右的顺序打印. 例如输入 8    /  \   6    10  /\     /\ 5  7   9  11 输出8    ...

  8. 8.python笔记之面向对象基础

    title: 8.Python笔记之面向对象基础 date: 2016-02-21 15:10:35 tags: Python categories: Python --- 面向对象思维导图 (来自1 ...

  9. ext树表

    aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAAA2UAAAHwCAIAAACpIFDdAAAgAElEQVR4nOy9f5Qb5ZnvWWQZlnO5Oc ...

  10. ubuntu 安装zabbix_agent端

    root@(none):~# apt-get install zabbix-agent root@(none):~# vi /etc/zabbix/zabbix_agentd.conf Server= ...