专栏首页mlHDUOJ---Piggy-Bank

HDUOJ---Piggy-Bank

Piggy-Bank

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9768    Accepted Submission(s): 4911

Problem Description

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! 

Input

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. 

Output

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.". 

Sample Input

3 10 110 2 1 1 30 50 10 110 2 1 1 50 30 1 6 2 10 3 20 4

Sample Output

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.

Source

Central Europe 1999

完全背包

代码:

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<stdlib.h>
 4 const int maxn=10005;
 5 const int inf=-0x3f3f3f3f;
 6 int dp[maxn],coin[501],weight[501];
 7 int main()
 8 {
 9     int i,j,n;
10     int test,low,high,cnt;
11     scanf("%d",&test);
12     while(test--)
13     {
14         scanf("%d%d",&low,&high);
15         cnt=high-low;
16         scanf("%d",&n);
17         for(i=0;i<n;i++)
18             scanf("%d%d",&coin[i],&weight[i]);
19         /*for(i=1;i<maxn;i++)
20             dp[i]=inf;*/
21         memset(dp,-1,sizeof(dp));
22         dp[0]=0;
23         for(i=0;i<n;i++)
24         {
25             for(j=weight[i];j<=cnt;j++)
26             {
27                 if(dp[j-weight[i]]>-1&&(dp[j]==-1||dp[j]>(dp[j-weight[i]]+coin[i])))
28                     dp[j]=dp[j-weight[i]]+coin[i];
29             }
30         }
31         if(dp[cnt]==-1)
32             printf("This is impossible.\n");
33         else
34             printf("The minimum amount of money in the piggy-bank is %d.\n",dp[cnt]);
35     }
36     return 0;
37 }

本文参与腾讯云自媒体分享计划,欢迎正在阅读的你也加入,一起分享。

我来说两句

0 条评论
登录 后参与评论

相关文章

  • POJ----The Suspects

    The Suspects Time Limit: 1000MS Memory Limit: 20000K Total Submissions: ...

    Gxjun
  • uva----(10794) A Different Task

     A Different Task  The (Three peg) Tower of Hanoi problem is a popular ...

    Gxjun
  • HDUOJ----专题训练

    Problem B Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (...

    Gxjun
  • Codeforces 839D Winter is here【数学:容斥原理】

    D. Winter is here time limit per test:3 seconds memory limit per test:256 megaby...

    Angel_Kitty
  • 4.1 数值积分、高等函数绘制

    is defined informally as the signed area of the region in the xy-plane that is ...

    周星星9527
  • CodeForces 639 A

    Bear and Displayed Friends time limit per test2 seconds memory limit per tes...

    ShenduCC
  • 非局域电学的Treftz函数 (CS)

    溶剂中的静电相互作用在生物物理系统中起着重要作用。文献中有一个共识,即水溶液的介电响应是非局部性的:极化不仅取决于给定点的电场,而且还取决于该点附近的电场。这通...

    管欣8078776
  • matplotlib.collections、(二)

    版权声明:本文为博主原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接和本声明。 ...

    于小勇
  • SAP S/4 HANA新变化-SD销售与分销

    1、数据库表变化 简化了数据模型: 取消了状态表VBUK, VBUP,状态表相关字段移到了销售对象表中,包括VBAK 、VBAP、LIKP、 LIPS、VBR...

    SAP最佳业务实践
  • 【Codeforces 738D】Sea Battle(贪心)

    http://codeforces.com/contest/738/problem/D

    饶文津

扫码关注云+社区

领取腾讯云代金券