poj1276Cash Machine【多重背包模板题】
来源:互联网 发布:安卓java模拟器apk 编辑:程序博客网 时间:2024/06/06 10:24
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.
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.
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.
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.
不解释,注意数组大小,和
hdu2191悼念512汶川大地震遇难同胞——珍惜现在,感恩生活【多重背包模板题】
一模一样==
#include <iostream>#include<cstdio>#include<cstring>using namespace std;int dp[100009],c[100009],w[100009],m[100009];int n,M,t;void zeropack(int cost,int weight){ for(int i=M;i>=cost;i--) dp[i]=max(dp[i-cost]+weight,dp[i]);}void completepack(int cost,int weight){ for(int i=cost;i<=M;i++) dp[i]=max(dp[i-cost]+weight,dp[i]);}void multipack(int cost,int weight,int num){ if(num*cost>=M) { completepack(cost,weight); return; } int k=1; while(k<num) { zeropack(k*cost,k*weight); num-=k; k*=2; } zeropack(cost*num,weight*num);}int main(){ //freopen("cin.txt","r",stdin); while(~scanf("%d%d",&M,&n)) { for(int i=1;i<=n;i++){scanf("%d%d",&m[i],&c[i]);w[i]=c[i];} memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) { multipack(c[i],w[i],m[i]); } printf("%d\n",dp[M]); } return 0;}
0 0
- poj1276Cash Machine【多重背包模板题】
- poj1276Cash Machine(多重背包)
- poj1276Cash Machine(多重背包)
- poj1276Cash Machine ——多重背包二进制优化
- POJ - 1276 Cash Machine 多重背包模板题
- PKU-1276-Cash Machine(多重背包模板)
- Cash Machine--多重背包
- Cash Machine (多重背包)
- Cash Machine -----多重背包
- Cash Machine --多重背包
- hdu1171(多重背包模板题)
- Dividing-多重背包模板题
- 多重背包模板题 背包问题V2
- hdu 2844 多重背包模板题 01背包、完全背包、多重背包模板
- Cash Machine----POJ_1276----多重背包
- poj1276(Cash Machine + 多重背包)
- POJ1276:Cash Machine(多重背包)
- POJ1276:Cash Machine(多重背包)
- 寻找和为定值的多个数
- 释放LINUX CACHEC的 小脚本
- 在Win10上发布Meteor应用
- Maven学习之17使用jetty来发布webapps(成功)
- Android DES MD5 UTF—8 BASE64 加密解密
- poj1276Cash Machine【多重背包模板题】
- 设计模式总结
- ListBox 控件
- nodejs __dirname 与 process.cwd(); 的区别
- 采样定理说的是什么?
- 过滤器与拦截器的区别
- 如果天生没什么运气,那你只能努力坚持这10个习惯了
- MSSQL通过SQL语句实现发邮件
- Android基础知识之智能指针:强指针和弱指针