poj 1787 背包+記錄路徑
阿新 • • 發佈:2017-11-03
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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.
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.
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.".
http://poj.org/problem?id=1787
Charlie‘s ChangeTime 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
Output
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