PAT1068. Find More Coins

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she must pay the exact amount. Since she has as many as 10⁴ coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find some coins to pay for it.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (<=10⁴, the total number of coins) and M(<=10², the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the face values V₁ <= V₂ <= ... <= V_k such that V₁ + V₂ + ... + V_k = M. All the numbers must be separated by a space, and there must be no extra space at the end of the line. If such a solution is not unique, output the smallest sequence. If there is no solution, output "No Solution" instead.

Note: sequence {A[1], A[2], ...} is said to be "smaller" than sequence {B[1], B[2], ...} if there exists k >= 1 such that A[i]=B[i] for all i < k, and A[k] < B[k].

8 9
5 9 8 7 2 3 4 1

1 3 5

4 8
7 2 4 3

No Solution

思路：典型的0-1背包动态规划，但是此题有一点需要注意，按照字典顺序进行排序，则需要将value进行从大到小排序，之所以从大到小排序，是因为利用dp一维数组进行的。应抓住此处要领。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <vector>
 6 using namespace std;
 7 #define MAX 10010
 8 int dp[MAX]={
 9     0
10 };
11 int value[MAX];
12 bool choice[MAX][MAX]={
13     false
14 };
15 bool cmp(int a,int b)
16 {
17     return a>b;
18 }
19 int main(int argc, char *argv[])
20 {
21     int N,M;
22     scanf("%d%d",&N,&M);
23     for(int i=1;i<=N;i++)
24        scanf("%d",&value[i]);
25     sort(value+1,value+N+1,cmp);  //排序的是地址 
26     for(int i=1;i<=N;i++)
27     {
28         for(int v=M;v>=value[i];v--)
29         {
30             if(dp[v]<=dp[v-value[i]]+value[i])
31             {
32                 choice[i][v]=true;
33                 dp[v]=dp[v-value[i]]+value[i];                
34             }
35         }
36     }
37     vector<int>ans;
38    // cout<<dp[N][M]<<endl;
39     if(dp[M]!=M)
40     {
41         printf("No Solution\n");
42     }
43     else
44     {
45         int k=N;
46         int num=0;
47         while(k>0)
48         {      
49             if(choice[k][M])
50             {
51                 M-=value[k];
52                 ans.push_back(value[k]);
53                 num++;
54             }
55             k--;
56         }
57         for(int i=0;i<ans.size();i++)
58         {
59             printf("%d",ans[i]);
60             if(i!=ans.size()-1)
61                putchar(‘ ‘);
62         }
63         putchar(‘\n‘);
64     }
65     return 0;
66 }

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