ural 1244. Gentlemen

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

inputoutput
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。
  1 #include <cstdio>
  2 #include <cstring>
  3 #include <cstdlib>
  4 #include <cmath>
  5 #include <deque>
  6 #include <vector>
  7 #include <queue>
  8 #include <iostream>
  9 #include <algorithm>
 10 #include <map>
 11 #include <set>
 12 #include <ctime>
 13 #include <iomanip>
 14 using namespace std;
 15 typedef long long LL;
 16 typedef double DB;
 17 #define For(i, s, t) for(int i = (s); i <= (t); i++)
 18 #define Ford(i, s, t) for(int i = (s); i >= (t); i--)
 19 #define Rep(i, t) for(int i = (0); i < (t); i++)
 20 #define Repn(i, t) for(int i = ((t)-1); i >= (0); i--)
 21 #define rep(i, x, t) for(int i = (x); i < (t); i++)
 22 #define MIT (2147483647)
 23 #define INF (1000000001)
 24 #define MLL (1000000000000000001LL)
 25 #define sz(x) ((int) (x).size())
 26 #define clr(x, y) memset(x, y, sizeof(x))
 27 #define puf push_front
 28 #define pub push_back
 29 #define pof pop_front
 30 #define pob pop_back
 31 #define ft first
 32 #define sd second
 33 #define mk make_pair
 34 inline void SetIO(string Name)
 35 {
 36     string Input = Name+".in",
 37     Output = Name+".out";
 38     freopen(Input.c_str(), "r", stdin),
 39     freopen(Output.c_str(), "w", stdout);
 40 }
 41 
 42 inline int Getint()
 43 {
 44     int Ret = 0;
 45     char Ch = ' ';
 46     bool Flag = 0;
 47     while(!(Ch >= '0' && Ch <= '9'))
 48     {
 49         if(Ch == '-') Flag ^= 1;
 50         Ch = getchar();
 51     }
 52     while(Ch >= '0' && Ch <= '9')
 53     {
 54         Ret = Ret * 10 + Ch - '0';
 55         Ch = getchar();
 56     }
 57     return Flag ? -Ret : Ret;
 58 }
 59 
 60 const int N = 110, M = 100010;
 61 int n, m, Arr[N];
 62 int Dp[M], G[M];
 63 bool Visit[N];
 64 vector<int> Ans;
 65 
 66 inline void Input() 
 67 {
 68     scanf("%d", &m);
 69     scanf("%d", &n);
 70     For(i, 1, n) scanf("%d", Arr + i);
 71 }
 72 
 73 inline void Search(int x)
 74 {
 75     if(!x) return;
 76     Visit[G[x]] = 1;
 77     Search(x - Arr[G[x]]);
 78 }
 79 
 80 inline void Solve()
 81 {
 82     Dp[0] = 1;
 83     int Max = 0;
 84     For(i, 1, n)
 85     {
 86         Max += Arr[i];
 87         if(Max > m) Max = m;
 88         Ford(j, Max, Arr[i])
 89             if(Dp[j - Arr[i]])
 90             {
 91                 Dp[j] += Dp[j - Arr[i]];
 92                 if(!G[j]) G[j] = i;
 93             }
 94     }
 95     
 96     if(Dp[m] > 1) puts("-1");
 97     else if(!Dp[m]) puts("0");
 98     else
 99     {
100         Search(m);
101         For(i, 1, n)
102             if(!Visit[i]) Ans.pub(i);
103         int Length = sz(Ans);
104         Rep(i, Length - 1) printf("%d ", Ans[i]);
105         printf("%d\n", Ans.back());
106     }
107 }
108 
109 int main()
110 {
111     #ifndef ONLINE_JUDGE 
112     SetIO("C");
113     #endif 
114     Input();
115     Solve();
116     return 0;
117 }
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posted @ 2015-11-14 16:03  yanzx6  阅读(150)  评论(0)  编辑  收藏  举报