HDU 5527---Too Rich（貪心+搜索）

Problem Description

You are a rich person, and you think your wallet is too heavy and full now. So you want to give me some money by buying a lovely pusheen sticker which costs pdollars from me. To make your wallet lighter, you decide to pay exactly p dollars by as many coins and/or banknotes as possible.

For example, if p=17 and you have two $10 coins, four $5 coins, and eight $1 coins, you will pay it by two $5 coins and seven $1 coins. But this task is incredibly hard since you are too rich and the sticker is too expensive and pusheen is too lovely, please write a program to calculate the best solution.

Input

The first line contains an integer T indicating the total number of test cases. Each test case is a line with 11 integers p,c1,c5,c10,c20,c50,c100,c200,c500,c1000,c2000, specifying the price of the pusheen sticker, and the number of coins and banknotes in each denomination. The number ci means how many coins/banknotes in denominations of i dollars in your wallet.

1≤T≤20000

0≤p≤109

0≤ci≤100000

Output

For each test case, please output the maximum number of coins and/or banknotes he can pay for exactly p dollars in a line. If you cannot pay for exactly p dollars, please simply output "-1".

Sample Input

3

17 8 4 2 0 0 0 0 0 0 0

100 99 0 0 0 0 0 0 0 0 0

2015 9 8 7 6 5 4 3 2 1 0

Sample Output

9

-1

36

題意：給了 p 表示要付的錢數，一個數列v[10],分別表示 1 ，5，10，20，50，100，200，500，1000，2000 元的錢幣數量，求用盡量多的錢幣剛好付清 p 元，輸出錢幣數。

思路：貪心，盡量用面值小的錢幣去籌，但是很可能面值小的錢幣不夠，所以從大面值開始考慮。初始化一個前綴和sum[12]，sum[i]表示v[1]~v[i]面值的錢幣和，tmp=rest-sum[i-1]，表示當前面值的錢幣應該付多少，cn=tmp/v[i] ，即表示當前面值的錢幣應該拿出多少張，如果tmp%v[i]！=0 ，那麼cn++，因為小於v[i]的錢幣無法籌出足夠的錢；另外要對於P=50 錢幣為 20,20,20，50 時，按照貪心策略3張20為60，所以不會取50，但是用3張20 無法籌出50元，所以必須每張面值的錢幣應該多考慮一張，比如對於這樣的數據：

p=1020 0 0 0 49 1 0 0 0 1 0；

代碼如下：

#include
#include
#include
#include
using namespace std;
int v[12]={0,1,5,10,20,50,100,200,500,1000,2000};
int c[12],sum[12];
int p,ans;
void dfs(int i,int rest,int count)
{
if(rest<0) return ; if(i==0) { if(rest==0) ans=max(ans,count); return ; } int tmp=max(0,rest-sum[i-1]); int cn=tmp/v[i]+(tmp%v[i]!=0); if(cn<=c[i]) dfs(i-1,rest-cn*v[i],count+cn); cn++; if(cn<=c[i]) dfs(i-1,rest-cn*v[i],count+cn); } int main { ///cout << "Hello world!" << endl; int T; cin>>T;
while(T--)
{
scanf("%d",&p);
for(int i=1;i<=10;i++) scanf("%d",&c[i]); sum[0]=0; for(int i=1;i<=10;i++) sum[i]=sum[i-1]+v[i]*c[i]; ans=-1; dfs(10,p,0); printf("%d ",ans); } return 0; } ///1020 0 0 0 49 1 0 0 0 1 0

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