HDOJ 1114
来源:互联网 发布:淘宝优惠券内容怎么写 编辑:程序博客网 时间:2024/06/16 11:58
Piggy-Bank
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8637 Accepted Submission(s): 4364
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!
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
310 11021 130 5010 11021 150 301 6210 320 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
Recommend
Eddy
简单的完全背包
#include <cstdio>#include <cmath>#include <algorithm>#include <iostream>#include <cstring>#include <map>#include <string>#include <stack>#include <cctype>#include <vector>#include <queue>#include <set>#include <iomanip>using namespace std;//#define Online_Judge#define outstars cout << "***********************" << endl;#define clr(a,b) memset(a,b,sizeof(a))#define lson l , mid , rt << 1#define rson mid + 1 , r , rt << 1|1#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)#define mk make_pairconst int inf = 1 << 30;const int MAXN = 10000 + 50;const int maxw = 100 + 20;const int MAXNNODE = 1000 +10;const long long LLMAX = 0x7fffffffffffffffLL;const long long LLMIN = 0x8000000000000000LL;const int INF = 0x7fffffff;const int IMIN = 0x80000000;#define eps 1e-8#define mod 10007typedef long long LL;const double PI = acos(-1.0);typedef double D;typedef pair<int , int> pii;const D e = 2.718281828459;int dp[MAXN];int w[550] , v[550];int main(){ //ios::sync_with_stdio(false);#ifdef Online_Judge freopen("in.txt","r",stdin); freopen("out.txt","w",stdout);#endif // Online_Judge int T; cin >> T; while(T--) { int n , m ; int a , b; scanf("%d%d" , &a , &b); m = b - a; scanf("%d" , &n); FOR(i , 0 , n)scanf("%d%d" , &v[i] , &w[i]); FORR(i , 0 , m )dp[i] = inf; dp[0] = 0; FOR(i , 0 , n) { FORR(j , w[i] , m) { dp[j] = min(dp[j] , dp[j - w[i]] + v[i]); } } if(dp[m] == inf)printf("This is impossible.\n"); else printf("The minimum amount of money in the piggy-bank is %d.\n" , dp[m]); } return 0;}
- HDOJ 1114
- HDOJ 1114 Piggy-Bank
- Piggy-Bank hdoj 1114
- hdoj 1114 Piggy-Bank
- HDOJ 1114 Piggy-Bank
- 【hdoj 1114】Piggy-Bank
- HDOJ-1114 Piggy-Bank
- HDOJ -- 1114 Piggy-Bank
- HDOJ
- hdoj
- hdoj
- HDOJ
- hdu/hdoj 1114 Piggy-Bank
- 完全背包模板 hdoj 1114
- HDOJ HDU 1114 Piggy-Bank
- zju 2014/ hdoj 1114-Piggy-Bank
- 完全背包——HDOJ 1114
- HDOJ 1114 Piggy-Bank 完全背包
- hdu 4502 吉哥系列故事——临时工计划 dp
- C++拷贝构造函数(深拷贝,浅拷贝)
- ubuntu截图
- [有图有真相]测试activity的生命周期
- C++弹出关闭光驱
- HDOJ 1114
- Ubuntu_Android SDK更新以及ADT更新出现问题的解决办法
- 编程之美——3.1 字符串移位包含的问题
- 反思 苦恼~
- android怎样调用@hide和internal API
- 转载: PostgreSQL学习手册(数据库维护)--恢复磁盘空间
- 公司估值(贴现现金流量法DCF)
- Application的学习总结
- hdu 1124 数学