POJ 1787 Charlie‘s Change
Description
Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at coffee vending machines at motorests. Charlie hates change. That is basically the setup of your next task.
Your program will be given numbers and types of coins Charlie has and
the coffee price. The coffee vending machines accept coins of values 1,
5, 10, and 25 cents. The program should output which coins Charlie has
to use paying the coffee so that he uses as
many coins as possible. Because Charlie really does not want any change
back he wants to pay the price exactly.
Input
Each line of the input
contains five integer numbers separated by a single space describing one
situation to solve. The first integer on the line P, 1 <= P <= 10
000, is the coffee price in cents. Next four integers, C1, C2, C3,
C4, 0 <= Ci <= 10 000, are the numbers of cents, nickels (5
cents), dimes (10 cents), and quarters (25 cents) in Charlie‘s valet.
The last line of the input contains five zeros and no output should be
generated for it.
Output
For each situation, your
program should output one line containing the string "Throw in T1 cents,
T2 nickels, T3 dimes, and T4 quarters.", where T1, T2, T3, T4 are the
numbers of coins of appropriate values Charlie should use to
pay the coffee while using as many coins as possible. In the case
Charlie does not possess enough change to pay the price of the coffee
exactly, your program should output "Charlie cannot buy coffee.".
Sample Input
12 5 3 1 2 16 0 0 0 1 0 0 0 0 0
Sample Output
Throw in 2 cents, 2 nickels, 0 dimes, and 0 quarters. Charlie cannot buy coffee. 题意:给你一个物品的价值 n 然后给你 4种硬币的数量（１ ５ １０ ２５这四种硬币） 问最多用多少硬币能正好凑够n 一眼看去就是多重背包还原路径这也是第一种解法
1 //#include<bits/stdc++.h>
2 #include<algorithm>
3 #include<iostream>
4 #include<cstring>
5 #include<cstdio>
6 using namespace std;
7 #define inf 0x3f3f3f3f
8 int vis[10001][90];
9 int dp[10001];
10 int q[4];
11 int main()
12 { freopen("in.txt","r",stdin);
13 ios::sync_with_stdio(false);
14 int n,c[4];
15 int v[4]={1,5,10,25};
16 while(cin>>n>>c[0]>>c[1]>>c[2]>>c[3]&&(n+c[1]+c[2]+c[3]+c[0]))
17 {
18 int a[101];
19 int b[101];
20 int d[101];
21 int cnt=0;
22 for(int i=0;i<4;i++)
23 {
24 int j=1;
25 while(j<=c[i])
26 {
27 a[cnt]=j*v[i];
28 b[cnt]=j;
29 d[cnt]=i;
30 cnt++;
31 c[i]-=j;
32 j*=2;
33 }
34 if(c[i])
35 {
36 a[cnt]=c[i]*v[i];
37 b[cnt]=c[i];
38 d[cnt]=i;
39 cnt++;
40 }
41
42
43 }
44
45 dp[0]=1;
46 for(int i=0;i<cnt;i++)
47 {
48
49 for(int j=n;j>=a[i];j--)
50 {
51 if(dp[j-a[i]]&&dp[j]<dp[j-a[i]]+b[i])
52 {
53 dp[j]=dp[j-a[i]]+b[i];
54 vis[j][i]=1;
55 }
56
57 }
58
59 }
60 // cout<<dp[n]<<endl;
61 if(!dp[n])
62 {
63 cout<<"Charlie cannot buy coffee."<<endl;
64 }
65 else
66 {
67 int j=n;
68 for(int i=cnt-1;i>=0;i--)
69 {
70 if(vis[j][i])
71 { // cout<<d[i]<<endl;
72 q[d[i]]+=b[i];
73 j-=a[i];
74 }
75
76 }
77 cout<<"Throw in "<<q[0]<<" cents, "<<q[1]<<" nickels, "<<q[2]<<" dimes, and "<<q[3]<<" quarters."<<endl;
78 }
79 for(int i=0;i<=n;i++)
80 {
81 dp[i]=0;
82 for(int j=0;j<cnt;j++)
83 vis[i][j]=0;
84 }
85 q[0]=q[1]=q[2]=q[3]=0;
86
87
88
89 }
90
91
92
93 return 0;
94 }
第二种完全背包 记录使用次数 大于时不更新
写的有点乱，就这吧，这种方法时间短（自己还是想不起来．．）
1 //#include<bits/stdc++.h>
2 #include<algorithm>
3 #include<iostream>
4 #include<cstring>
5 #include<cstdio>
6 using namespace std;
7 #define inf 0x3f3f3f3f
8 int dp[10001];
9 int q[105];
10 int fa[10001];
11 int usr[10001];
12 int main()
13 { //freopen("in.txt","r",stdin);
14 ios::sync_with_stdio(false);
15 int n,c[6];
16 int v[9]={0,1,5,10,25};
17 while(cin>>n>>c[1]>>c[2]>>c[3]>>c[4]&&(n+c[1]+c[2]+c[3]+c[4]))
18 {
19 memset(fa,0,sizeof(fa));
20 memset(dp,0,sizeof(dp));
21 memset(q,0,sizeof(q));
22 dp[0]=1;
23 for(int i=1;i<=4;i++)
24 {
25 memset(usr,0,sizeof(usr));
26 for(int j=v[i];j<=n;j++)
27 {
28
29 if(dp[j-v[i]]&&dp[j]<dp[j-v[i]]+1&&usr[j-v[i]]<c[i])
30 {
31 dp[j]=dp[j-v[i]]+1;
32 usr[j]=usr[j-v[i]]+1;
33 fa[j]=j-v[i];
34 }
35
36 }
37
38 }
39 if(!dp[n])
40 {
41 cout<<"Charlie cannot buy coffee."<<endl;
42 }
43 else
44 {
45 while(n!=0)
46 {
47 q[n-fa[n]]++;
48 n=fa[n];
49
50 }
51 cout<<"Throw in "<<q[1]<<" cents, "<<q[5]<<" nickels, "<<q[10]<<" dimes, and "<<q[25]<<" quarters."<<endl;
52 }
53
54 //q[0]=q[1]=q[2]=q[3]=0;
55
56
57
58 }
59
60
61
62 return 0;
63 }
64 /*
65 #include<bits/stdc++.h>
66 #include<iostream>
67 #include<cstring>
68 using namespace std;
69 const int N = 10005;
70
71 struct ac{
72 int w, v, c;
73 }a[N*20];
74
75 int dp[N], w[8] = {0,1,5,10,25};
76 int path[60][N], ans[105];
77 int main()
78 { freopen("in.txt","r",stdin);
79 cin.sync_with_stdio(false);
80 int p,c[14];
81 while(cin >> p >> c[1] >> c[2] >> c[3] >> c[4])
82 {
83 if(p+c[1]+c[2]+c[3]+c[4] == 0) break;
84 int cnt = 0;
85 for(int i = 1; i <= 4; i++)
86 {
87 for( int j = 1; j <= c[i]; j <<= 1)
88 {
89 a[++cnt].w = j*w[i];
90 a[cnt].c = j;
91 a[cnt].v = w[i];
92 c[i] -= j;
93 }
94
95 if(c[i])
96 {
97 a[++cnt].w = w[i]*c[i];
98 a[cnt].c = c[i];
99 a[cnt].v = w[i];
100 }
101 }
102 memset(dp, 0, sizeof(dp));
103 memset(ans, 0, sizeof(ans));
104 for(int i = 1; i <= cnt; i++)
105 {
106 for(int j = 1; j <= p; j++) path[i][j] = 0;
107 }
108 cout<<cnt<<endl;
109 dp[0] = 1;
110 for(int i = 1; i <= cnt; i++)
111 {
112 for(int j = p; j >= a[i].w; j--)
113 {
114 if(dp[j-a[i].w] && dp[j] < dp[j-a[i].w] + a[i].c)
115 {
116 dp[j] = dp[j-a[i].w] + a[i].c;
117 path[i][j] = 1;
118 }
119 }
120 }
121 //cout << dp[p] << endl;
122
123 if(dp[p])
124 {
125 for(int i = cnt; i >= 1; i--)
126 {
127 if(path[i][p])
128 {
129 ans[a[i].v] += a[i].c;
130 p -= a[i].w;
131 }
132 }
133 cout << "Throw in " << ans[1] << " cents, " << ans[5] << " nickels, " << ans[10] << " dimes, and " << ans[25] << " quarters." << endl;
134 }
135 else cout << "Charlie cannot buy coffee.\n";
136
137 }
138 return 0;
139 }
140 */
原文地址：https://www.cnblogs.com/qq1976648487/p/8407199.html