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http://poj.org/problem?id=1787

Charlie‘s Change

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 4512		Accepted: 1425

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.

Source

CTU Open 2003 給出了四種不同面值的硬幣及其數量，問組成P價值所用的最多數量的硬幣是多少，並輸出這個方案。應該是多重背包把這個，用二進制優化個數之後就不會T了，剛開始卡了很久因為兩層循環我寫反了，一直沒寫過背包有點蒙，由於用到了一維數組的優化，所以對於當前的某件物品，當前的價值用到的子問題必須不涉及到這件物品。

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 #define inf 0x3f3f3f3f
 6 int coin[5]={0,1,5,10,25};
 7 int f[10005];
 8 int book[5];
 9 struct date
10 {
11     int num,type;
12 }Q[10005];
13 int main()
14 {
15     int C[5],P,n,m,i,j,k;
16     while(cin>>P>>C[1]>>C[2]>>C[3]>>C[4]){
17         if(!(P+C[1]+C[2]+C[3]+C[4])) break;
18         memset(f,-inf,sizeof(f));
19         memset(Q,0,sizeof(Q));
20         memset(book,0,sizeof(book));
21         int W,L,l=0;
22         f[0]=0;
23         for(i=1;i<=4;++i)
24         {
25             l=0;
26             for(k=1;C[i];C[i]-=k,k*=2){
27                 if(C[i]<k) k=C[i];
28                // cout<<i<<‘ ‘<<k<<endl;
29                 W=k*coin[i];
30             for(j=P;j>=W;--j)
31             {
32 
33 
34                     if(f[j-W]!=-inf&&f[j]<f[j-W]+k)
35                         {
36                             f[j]=f[j-W]+k;
37                             Q[j].num=k;
38                             Q[j].type=i;
39                         }
40             }
41             }
42         }//puts("dd");
43        // cout<<f[P]<<endl;
44        if(f[P]<0) puts("Charlie cannot buy coffee.");
45        else{
46             j=P;
47             i=0;
48             while(j){
49                 book[Q[j].type]+=Q[j].num;
50                 j-=coin[Q[j].type]*Q[j].num;
51             }
52             printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",book[1],book[2],book[3],book[4]);
53        }
54     }
55     return 0;
56 }

poj 1787 背包+記錄路徑