poj1276--Cash Machine(多重背包判可达)

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. 

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. 

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

735 3  4 125  6 5  3 350
633 4  500 30  6 100  1 5  0 1
735 0
0 3  10 100  10 50  10 10

735
630
0
0

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.

给出n种货币，每种的价值和数量，问可得到最接近case的是多少？

多重背包问题，使用二进制转化为01背包。

从m转化为 12,4,2^(k-1) , m-2^k+1 ; k是由m-2^k+1>0的最大整数得到

 

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int dp[110000] ;
int m[20] , a[20] ;
int main()
{
    int c , n , i , j , l , k , mm ;
    while(scanf("%d %d", &c, &n)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        dp[0] = 1 ;
        for(i = 0 ; i < n ; i++)
            scanf("%d %d", &m[i], &a[i]) ;
        for(i = 0 ; i < n ; i++)
        {
            k = 0 ;
            while( (1<<k) <= m[i] )
                k++ ;
            k-- ;
            if(k == -1) continue ;
            mm = (m[i] - (1<<k) +1  ) * a[i] ;
            for(j = c ; j-mm >= 0 ; j--)
                dp[j] += dp[j-mm] ;
            for(l = 0 ; l < k ; l++)
            {
                mm = (1<<l)*a[i] ;
                for(j = c ; j-mm >= 0 ; j--)
                    dp[j] += dp[j-mm] ;
            }
        }
        for(i = c ; i >= 0 ; i--)
            if( dp[i] )
                break;
        printf("%d\n", i);
    }
    return 0;
}