Cash Machine poj 1276

Cash Machine

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 33160 Accepted: 12015

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.

/*

有各种不同面值的货币，每种面值的货币有不同的数量，请找出利用这些货币可以凑成的最接近且小于等于给定的数字cash的金额。

*/

思想就是用多重背包的思想，我们可以转化成数组的形式，只要是能凑出的钱数就使value[钱数] = 1,只要找到最近的1就行了。

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;int value[100010];struct beibao{    int v;    int b;} a[20];int main(){    int i, j, k;    int m, n;    while ( ~scanf ( "%d %d", &n, &m ) )    {        for(  i = 1; i <= m; i++ )            scanf ( "%d %d", &a[i].b , &a[i].v );        if(n == 0||m == 0)          {              printf("0\n");              continue;          }          memset(value, 0, sizeof(value) );        value[0] = 1;        int money = 0;        int sum;        for(  i = 1; i <= m; i++)        {            for ( j = money;j >= 0; j-- )            {                if ( value[j] )                {                    for ( k = 1; k <= a[i].b; k++ )                    {                        sum = j+a[i].v*k;                        if ( sum > n )                            break;                        value[sum] = 1;                        if( sum > money )                            money = sum;                    }                }            }        }        for ( i = n;i >= 0; i-- )        {            if ( value[i] != 0 )                break;        }        printf ( "%d\n", i );    }    return 0;}

代码菜鸟，如有错误，请多包涵！！！

如有帮助记得支持我一下，谢谢！！！