uva 10306 - e-Coins(二维完全背包)
来源:互联网 发布:js中json数组添加元素 编辑:程序博客网 时间:2024/06/18 14:16
1、http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1247
2、题目大意:
对于每个样例,先给定两个数n,m,分别表示有n种硬币,对于每一种硬币有两个价值,分别记做x,y,题目要求从中选择一些硬币,使得满足m=sqrt(x^2+y^2),其中是选出的硬币的所有x的和,y是所有选出的硬币的y的和,硬币有无数多个,现在要求的是,满足上述要求使用的最少的硬币数
本题可以看成是一个二维的完全背包问题,其中m看做是容量,个数看做是价值,现在转化成当前容量下用的最小的价值
3、AC代码:
#include<stdio.h>#define N 45#define INF 0x7fffffff#include<algorithm>using namespace std;int w1[N],w2[N];int dp[305][305];int main(){ int t,n,m; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); for(int i=0;i<n;i++) scanf("%d%d",&w1[i],&w2[i]); for(int i=0;i<=m;i++) { for(int j=0;j<=m;j++) dp[i][j]=INF; } dp[0][0]=0; for(int i=0;i<n;i++) { for(int j=w1[i];j<=m;j++) { for(int k=w2[i];k<=m;k++) { if(dp[j-w1[i]][k-w2[i]]!=INF) { dp[j][k]=min(dp[j][k],dp[j-w1[i]][k-w2[i]]+1); } } } } int ans=INF; int mm=m*m; for(int i=0;i<=m;i++) { for(int j=0;j<=m;j++) { if(i*i+j*j==mm && dp[i][j]!=INF) { ans=min(ans,dp[i][j]); } } } if(ans!=INF) printf("%d\n",ans); else printf("not possible\n"); } return 0;}
4、题目:
Problem G
e-Coins
Input: standard input
Output: standard output
Time Limit: 10 seconds
Memory Limit: 32 MB
At the Department for Bills and Coins, an extension of today's monetary system has newly been proposed, in order to make it fit the new economy better. A number of new so called e-coins will be produced, which, in addition to having a value in the normal sense of today, also have an InfoTechnological value. The goal of this reform is, of course, to make justice to the economy of numerous dotcom companies which, despite the fact that they are low on money surely have a lot of IT inside. All money of the old kind will keep its conventional value and get zero InfoTechnological value.
To successfully make value comparisons in the new system, something called the e-modulus is introduced. This is calculated asSQRT(X*X+Y*Y), where X and Y hold the sums of the conventional and InfoTechnological values respectively. For instance, money with a conventional value of $3 altogether and an InfoTechnological value of $4 will get an e-modulus of $5. Bear in mind that you have to calculate the sums of the conventional and InfoTechnological values separately before you calculate the e-modulus of the money.
To simplify the move to e-currency, you are assigned to write a program that, given the e-modulus that shall be reached and a list of the different types of e-coins that are available, calculates the smallest amount of e-coins that are needed to exactly match the e-modulus. There is no limit on how many e-coins of each type that may be used to match the given e-modulus.
Input
A line with the number of problems n (0<n<=100), followed by n times:
- A line with the integers m (0<m<=40) and S (0<S<=300), where m indicates the number of different e-coin types that exist in the problem, and S states the value of the e-modulus that shall be matched exactly.
- m lines, each consisting of one pair of non-negative integers describing the value of an e-coin. The first number in the pair states the conventional value, and the second number holds the InfoTechnological value of the coin.
When more than one number is present on a line, they will be separated by a space. Between each problem, there will be one blank line.
Output
The output consists of n lines. Each line contains either a single integer holding the number of coins necessary to reach the specified e-modulus Sor, if S cannot be reached, the string "not possible".
Sample Input:
3
2 5
0 2
2 0
3 20
0 2
2 0
2 1
3 5
3 0
0 4
5 5
Sample Output:
not possible
10
2
- uva 10306 e-Coins(二维完全背包)
- uva 10306 - e-Coins(二维完全背包)
- UVA 10306 e-Coins(二维完全背包)
- UVA 10306 e-Coins(二维完全背包)
- UVA 10306 - e-Coins(完全背包)
- uva 10306 - e-Coins(完全背包)
- UVA 10306 e-Coins(完全背包: 二维限制条件)
- UVA10306 - e-Coins(二维完全背包)
- UVA - 10306 e-Coins 二维的01背包问题
- UVA 10306 e-Coins 电子硬币 完全背包
- e-Coins - UVa 10306 dp背包
- uva_10306 - e-Coins (二维背包)
- uva10306 - e-Coins(完全背包)
- Coins(完全背包)
- uva 10306 e-Coins
- UVA 10306 e-Coins
- uva 10306 e-Coins
- UVa 10306 - e-Coins
- Halcon C#.net 编程指导(4): 可视化
- poj 2411 Mondriaan's Dream
- myeclipse不编译的解决办法
- Hama安装及示例运行
- JNDI
- uva 10306 - e-Coins(二维完全背包)
- 【2014校招】大众点评面试
- 【2014校招】京东网银在线笔试
- 如何处理婚姻问题
- ubuntu系统下更新目录下所有文件时间戳的方法
- strcpy()和memcpy()函数的区别
- 不借助第三方类库,使用JDK自带API操作XML
- Android 的main.mk完整分析
- Sublime Text 2破解