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. UIImage 和 iOS 图片压缩UIImage / UIImageVIew

    UIImageView 制作气泡 stretchableImageWithLeftCapWidth http://blog.csdn.net/justinjing0612/article/detail ...

  2. 影像工作站的数据库安装错误之Win7系统下pg服务无法启动

    1.关闭批处理 2.修改 PG安装路径下的Data文件下的pg_hba.conf文件中去掉IPv6的井号,如下图 3.结束pg进程 4.重启PG服务.

  3. poj2996 模拟

    Help Me with the Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3713   Accepted:  ...

  4. django admin 扩展

    添加自定义动作: 例子,添加一个方法,批量更新文章,代码如下: from django.contrib import admin from myapp.models import Article de ...

  5. Linux时间同步配置方法

    由于是在做mongoDB的实验中再一次的遇到了mongos路由节点同步时由于ntp时间的问题导致同步非常的慢.故写了个时间同步的语句===> while :; do rdate -s 192.1 ...

  6. Windows环境下的jekyll本地搭建

    一.配置ruby环境 由于jekyll是用ruby语言写的一个静态网页生成工具,所以要搭建jekyll本地环境就需要先配置好ruby环境. 1)去官网下载Ruby:https://www.ruby-l ...

  7. Unique Binary Search Trees I & II

    Given n, how many structurally unique BSTs (binary search trees) that store values 1...n? Example Gi ...

  8. 算法:comparable比较器的排序原理实现(二叉树中序排序)

    Comparable比较器排序远离实现 package test.java.api.api13; /** * 手工实现二叉树的比较算法: 第一遍感觉很神秘,但是真正自己写下来,就感觉很简单,理解就好: ...

  9. codeforces A. Vasily the Bear and Triangle 解题报告

    题目链接:http://codeforces.com/problemset/problem/336/A 好简单的一条数学题,是8月9日的.比赛中没有做出来,今天看,从pupil变成Newbie了,那个 ...

  10. 【USACO】milk3

    倒牛奶的问题, 开始看感觉跟倒水的问题很像, 想直接找规律, 写个类似于循环取余的代码. 但后来发现不行,因为这道题有三个桶,水量也是有限制的.只好用模拟的方法把所有的情况都试一遍. 建一个state ...