POJ 1276 Cash Machine (多重背包)

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

题意

一个取款机有N种钞票，每种钞票有nk张，面额为Dk，给定一个取款金额cash，可行的、不超过该金额的吐钞方案最大是多少钱？

思路

多重背包模版题目，具体查看 背包九讲 。

代码中加入了二进制优化。

AC 代码

#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;}