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POJ 1276 Cash Machine (多重背包)

2019-11-11 02:06:35
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Description

A Bank plans to install a machine for cash withdrawal. The machine is able to deliver apPRopriate @ bills for a requested cash amount. The machine uses exactly N distinct bill denominations, say Dk, k=1,N, and for each denomination Dk the machine has a supply of nk bills. For example,

N=3, n1=10, D1=100, n2=4, D2=50, n3=5, D3=10

means the machine has a supply of 10 bills of @100 each, 4 bills of @50 each, and 5 bills of @10 each.

Call cash the requested amount of cash the machine should deliver and write a program that computes the maximum amount of cash less than or equal to cash that can be effectively delivered according to the available bill supply of the machine.

Notes:

@ is the symbol of the currency delivered by the machine. For instance, @ may stand for dollar, euro, pound etc.

Input

The program input is from standard input. Each data set in the input stands for a particular transaction and has the format:

cash N n1 D1 n2 D2 … nN DN

where 0 <= cash <= 100000 is the amount of cash requested, 0 <=N <= 10 is the number of bill denominations and 0 <= nk <= 1000 is the number of available bills for the Dk denomination, 1 <= Dk <= 1000, k=1,N. White spaces can occur freely between the numbers in the input. The input data are correct.

Output

For each set of data the program prints the result to the standard output on a separate line as shown in the examples below.

Sample Input

735 3 4 125 6 5 3 350633 4 500 30 6 100 1 5 0 1735 00 3 10 100 10 50 10 10

Sample Output

73563000

題意

一個(gè)取款機(jī)有N種鈔票,每種鈔票有nk張,面額為Dk,給定一個(gè)取款金額cash,可行的、不超過(guò)該金額的吐鈔方案最大是多少錢(qián)?

思路

多重背包模版題目,具體查看 背包九講 。

代碼中加入了二進(jìn)制優(yōu)化。

AC 代碼

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>using namespace std;#include<vector>#include<queue>#define MAXX 110000int p[MAXX],h[MAXX],c[MAXX];int dp[MAXX];void solve(int pi,int hi,int n){ for(int i=n; i>=pi; i--) dp[i]=max(dp[i],dp[i-pi]+hi);}void mult(int pi,int hi,int ci,int n){ int k=1; while(k<=ci) { solve(k*pi,k*hi,n); ci-=k; k<<=1; } solve(ci*pi,ci*hi,n);}int main(){ int n,m; while(~scanf("%d%d",&n,&m)) { for(int i=0; i<m; i++) scanf("%d%d",c+i,p+i); memset(dp,0,sizeof(dp)); for(int i=0; i<m; i++) mult(p[i],p[i],c[i],n); printf("%d/n",dp[n]); } return 0;}
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