题目描述

In a 10-dollar shop, everything is worthy 10 dollars or less. In order to serve customers more effectively at the cashier, change needs to be provided in the minimum number of coins.
In this problem, you are going to provide a given value of the change in different coins. Write a program to calculate the number of coins needed for each type of coin.
The input includes a value v, a size of the coinage set n, and a face value of each coin, f1, f2, ..., fn. The output is a list of numbers, namely, c1, ..., cn, indicating the number of coins needed for each type of coin. There may be many ways for the change. The value v is an integer satisfying 0 < v ≤ 2000, representing the change required
in cents. The face value of a coin is less than or equal to 10000. The output of your program should take the combination with the least number of coins needed.
For example, the Hong Kong coinage issued by the Hong Kong Monetary Authority consists of 10 cents, 20 cents, 50 cents, 1 dollar, 2 dollars, 5 dollars and 10 dollars would be represented in the input by n = 7, f1 = 10, f2 = 20, f3 = 50, f4 = 100, f5 = 200, f6 = 500, f7 = 1000.

输入

The test data may contain many test cases, please process it to the end of the file.
Each test case contains integers v, n, f1, ..., fn in a line. It is guaranteed that n ≤ 10 and 0 < f1 < f2 < ...< fn.

输出

The output be n numbers in a line, separated by space. If there is no possible change, your output should be a single −1. If there are more than one possible solutions, your program should output the one that uses more coins of a lower face value.

样例输入

2000 7 10 20 50 100 200 500 1000
250 4 10 20 125 150
35 4 10 20 125 150
48 4 1 8 16 20
40 4 1 10 13 37
43 5 1 2 21 40 80

样例输出

0 0 0 0 0 0 2
0 0 2 0
-1
0 1 0 2
3 0 0 1
1 1 0 1 0
 1 #include<bits/stdc++.h>
 2 const int inf=0x3f3f3f;
 3 using namespace std;
 4 int dp[10010];
 5 int s[20];
 6 int f[20];
 7 int n;
 8 void ptans(int t)//递推找上一个最小的钱
 9 {
10     for(int i=1;i<=n;i++)
11     if(t>=s[i]&&dp[t]==dp[t-s[i]]+1){
12         f[i]++;
13         ptans(t-s[i]);
14         break;
15     }
16 }
17 int main()
18 {
19     int i,j;
20     int ans;
21     ios::sync_with_stdio(false);
22     while(cin>>ans){
23         memset(dp,0,sizeof(dp));
24         memset(s,0,sizeof(s));
25         memset(f,0,sizeof(f));//初始化
26         for(i=1;i<=ans;i++) dp[i]=inf;
27         cin>>n;
28         for(i=1;i<=n;i++) cin>>s[i];
29         for(i=1;i<=ans;i++)
30           for(j=1;j<=n;j++)  if(i>=s[j]) dp[i]=min(dp[i],dp[i-s[j]]+1);
31         if(dp[ans]==inf) cout<<-1<<endl;
32         else{
33         ptans(ans);
34         for(i=1;i<=n;i++){
35             if(i!=1) cout<<" ";
36             cout<<f[i];
37         }
38         cout<<endl;
39         }
40     }
41     return 0;
42 }
View Code

DAG 相当于一环套一环，但一个并不能直接或间接套在自己内部

就跟导弹拦截类似的那种

#include<bits/stdc++.h>
const int inf=0x3f3f3f;
using namespace std;
int dp[10010];
int s[20];
int f[20];
int gf[10010];
int n;
int main()
{
    int i,j;
    int ans;
    ios::sync_with_stdio(false);
    while(cin>>ans){
        memset(dp,0,sizeof(dp));
        memset(s,0,sizeof(s));
        memset(f,0,sizeof(f));
        memset(gf,0,sizeof(gf));
        for(i=1;i<=ans;i++) dp[i]=inf;
        cin>>n;
        for(i=1;i<=n;i++) cin>>s[i];
        for(i=1;i<=ans;i++)
            for(j=1;j<=n;j++)
        if(i>=s[j]){
            if(dp[i]>dp[i-s[j]]+1){
                dp[i]=dp[i-s[j]]+1;
                gf[i]=j;
            }
        }
        if(dp[ans]==inf) cout<<-1<<endl;
        else{
        int op=ans;
        while(op!=0){
            f[gf[op]]++;
            op=op-s[gf[op]];
        }
        for(i=1;i<=n;i++){
            if(i!=1) cout<<" ";
            cout<<f[i];
        }
        cout<<endl;
        }
    }
    return 0;
}
View Code

这个代码没有递归的过程 而是直接在更新的过程中记录路径