poj 1276 Cash Machine

Cash Machine

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 30361		Accepted: 10943

DescriptionA 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 supplyof 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. InputThe 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 numbersin the input. The input data are correct. OutputFor 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

多重背包问题

#include<stdio.h>#include<string.h>#include<algorithm>#include<math.h>using namespace std;int value[100010];int dp[100010];int n,cashmoney,cnt;int package(){int i,j;for(i=1;i<=cnt;i++)for(j=cashmoney;j>=value[i];j--){dp[j]=max(dp[j],dp[j-value[i]]+value[i]);}return dp[cashmoney];}int main(){while(scanf("%d%d",&cashmoney,&n)!=EOF){cnt=0;int i,j,s;for(i=1;i<=n;i++){int num,money;scanf("%d%d",&num,&money);for(j=1;j<=num;j*=2){value[++cnt]=money*j;num-=j;}if(num>0)value[++cnt]=num*money;}for(j=1;j<=cnt;j++)printf("%d ",value[j]);printf("\n");s=package();printf("%d\n",s);memset(value,0,sizeof(value));memset(dp,0,sizeof(dp));}return 0;}