九度OJ 1454 Piggy-Bank(完全背包)
来源:互联网 发布:jquery数组中移除元素 编辑:程序博客网 时间:2024/06/08 05:47
- 题目描述:
Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he takes all the coins and throws them into a piggy-bank. You know that this process is irreversible, the coins cannot be removed without breaking the pig. After a sufficiently long time, there should be enough cash in the piggy-bank to pay everything that needs to be paid.
But there is a big problem with piggy-banks. It is not possible to determine how much money is inside. So we might break the pig into pieces only to find out that there is not enough money. Clearly, we want to avoid this unpleasant situation. The only possibility is to weigh the piggy-bank and try to guess how many coins are inside. Assume that we are able to determine the weight of the pig exactly and that we know the weights of all coins of a given currency. Then there is some minimum amount of money in the piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. We need your help. No more prematurely broken pigs!
- 输入:
The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers E and F. They indicate the weight of an empty pig and of the pig filled with coins. Both weights are given in grams. No pig will weigh more than 10 kg, that means 1 <= E <= F <= 10000. On the second line of each test case, there is an integer number N (1 <= N <= 500) that gives the number of various coins used in the given currency. Following this are exactly N lines, each specifying one coin type. These lines contain two integers each, Pand W (1 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it's weight in grams.
- 输出:
Print exactly one line of output for each test case. The line must contain the sentence "The minimum amount of money in the piggy-bank is X." where X is the minimum amount of money that can be achieved using coins with the given total weight. If the weight cannot be reached exactly, print a line "This is impossible.".
- 样例输入:
310 11021 130 5010 11021 150 301 6210 320 4
- 样例输出:
The minimum amount of money in the piggy-bank is 60.The minimum amount of money in the piggy-bank is 100.This is impossible.
#include <stdio.h>const int INF = 0x7FFFFFFF;struct COIN{int P;int W;}coin[501];int min(int a, int b){return a < b ? a : b;}int main(){//freopen("TEST.txt","r",stdin);int T;scanf("%d",&T);while(T--){int E, F;scanf("%d%d",&E,&F);int Vol = F - E;int n;scanf("%d",&n);int i, j;for(i = 0; i < n; i++)scanf("%d%d",&coin[i].P,&coin[i].W);int dp[10001];dp[0] = 0;for(i = 1; i <= Vol; i++)dp[i] = INF;for(i = 0; i < n; i++){for(j = coin[i].W; j <= Vol; j++){if(dp[j-coin[i].W] != INF)dp[j] = min(dp[j-coin[i].W]+coin[i].P,dp[j]);}}if(dp[Vol] != INF)printf("The minimum amount of money in the piggy-bank is %d.\n",dp[Vol]);elseprintf("This is impossible.\n");}return 0;}
- 九度OJ 1454 Piggy-Bank(完全背包)
- 九度oj 题目1454:Piggy-Bank
- 完全背包- Piggy-Bank
- Piggy-Bank(完全背包)
- 完全背包,piggy-bank
- Piggy-Bank (完全背包)
- Piggy-Bank完全背包
- 【完全背包】Piggy - Bank
- Piggy-Bank(完全背包)
- Piggy-Bank(背包九讲_完全背包)
- hdu1114 Piggy-Bank 完全背包
- Piggy-Bank(完全背包)
- Piggy-Bank hdu1114 完全背包
- Piggy-Bank 完全背包问题
- HD Piggy-Bank完全背包
- hdu1114 Piggy-Bank (完全背包)
- hdu-Piggy-Bank(完全背包)
- poj1384 Piggy-Bank(完全背包)
- 经纬度坐标计算距离
- 员工总是请假,处理方法
- 怎样网购才更省钱
- Lua5.1中的API函数
- Linux如何查找文件安装路径
- 九度OJ 1454 Piggy-Bank(完全背包)
- Linux 命令行模式采用英文, 界面采用英文
- 排序算法之——桶排序
- 乐蜂网目标独立上市 唯品会向其派驻CEO、CFO
- 西部数据硬盘第一次启动很慢
- 比较MySQL 5.6与前版的同步协议
- struts2集成javamail发邮件(带附件)实践记录
- 免费的21个UI界面设计工具、资源及网站
- 科大讯飞发布四川话语音识别技术,号称识别准确率超过85%