Cash Machine

Problem 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: <br> <br>cash N n1 D1 n2 D2 ... nN DN <br> <br>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. <br>

 

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. <br>

 

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
多重背包模版题
每行数据第一个是要求价值 共有几件物品， 第一件物品的个数，第一件物品的价值.....
要求找到能得到的距离要求最近的价值（不能比要求价值大）
#include<iostream>#include<math.h>#include<stdio.h>using namespace std;int main(){    int n,m;    while(~scanf("%d %d",&n,&m))    {        int a[1000];        int t=0;        for(int i=0;i<m;i++)        {            int s1,v1;            scanf("%d %d",&s1,&v1);            for(int j=1;s1>0;j=j*2)            {                if(s1>=j)                    s1=s1-j,a[t]=j*v1,t++;                else                    a[t]=s1*v1,s1=0,t++;            }        }        int b[100010]={0};        int ma=0;        for(int i=0;i<t;i++)        {            for(int j=n;j>=a[i];j--)                b[j]=max(b[j],b[j-a[i]]+a[i]),ma=max(ma,b[j]);        }        printf("%d\n",ma);    }    return 0;}