POJ

Cash Machine

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

Hint

The first data set designates a transaction where the amount of cash requested is @735. The machine contains 3 bill denominations: 4 bills of @125, 6 bills of @5, and 3 bills of @350. The machine can deliver the exact amount of requested cash. 

In the second case the bill supply of the machine does not fit the exact amount of cash requested. The maximum cash that can be delivered is @630. Notice that there can be several possibilities to combine the bills in the machine for matching the delivered cash. 

In the third case the machine is empty and no cash is delivered. In the fourth case the amount of cash requested is @0 and, therefore, the machine delivers no cash.

题意：多重背包裸题

解题思路：多重背包裸题，套模板即可，由于多重背包用到了01背包和完全背包，所以这里直接结合三种背包写成一个模板。代码有详细注释。多重背包的解法是用了二进制的思想，把N个物品分解成logN件，他们可以组合出各种各样的物品。详见代码。

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <set>#include <vector>#include <map>#include <stack>#include <set>#include <algorithm>#include<queue>using namespace std;typedef long long ll;const int MAXN=2000005;//输入外挂inline int scanf(int &num){    char in;bool IsN=false;    in=getchar();    if(in==EOF) return -1;    while(in!='-'&&(in<'0'||in>'9')) in=getchar();    if(in=='-'){ IsN=true;num=0;}    else num=in-'0';    while(in=getchar(),in>='0'&&in<='9'){        num*=10,num+=in-'0';    }    if(IsN) num=-num;    return 1;}int W,N;int num[100005];int value[100005];int weight[100005];int dp[100005];//01背包模板void zero_one_pack(int w,int v){    for(int i=W;i>=w;i--)        dp[i]=max(dp[i-w]+v,dp[i]);}//完全背包模板void complete_pack(int w,int v){    for(int i=w;i<=W;i++)        dp[i]=max(dp[i-w]+v,dp[i]);}//多重背包模板void multi_pack(int w,int v,int n){    //如果该物品的数量*重量比背包容量还大，那么其实就是一个完全背包问题，直接转化为完全背包求解    if(n*w>=W){        complete_pack(w,v);        return;    }    //否则将该物品分解成几个小物品，然后用01背包去求解。详见背包九讲    int k=1;    while(k<n){        zero_one_pack(k*w,k*v);        n-=k;        k*=2;    }    zero_one_pack(n*w,n*v);    }int main(){    while(~scanf(W)){        scanf(N);        for(int i=1;i<=N;i++){            scanf(num[i]);            scanf(value[i]);            weight[i]=value[i];        }        memset(dp,0,sizeof(dp));        //对于每一个物品，用多重背包求解        for(int i=1;i<=N;i++)            multi_pack(weight[i],value[i],num[i]);        printf("%d\n",dp[W]);    }    return 0;}