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现测量出存钱罐初始的重量(E)和存满钱时的重量(F),问在这N种硬币中,如何选取,使得选取的硬币的总重量恰好等于(F-E),且总价值最小。 如果可以找到,输出"The minimum amount of money in the piggy-bank is num".(num : 总和最小的面值) 如果不能通过组合使得硬币总重量恰好等于(F-E),则输出"This is...
Capacity: 1000/2000/3000/5000/10000 euro,Accommodates a variety of savings goals, from 1000 to 10000 euros. Material: Wood,Crafted from premium wood, this piggy bank offers a natural, eco-friendly way to save. Customer Reviews (9)
1 <= E <= F <= 10000. 第二行包含了一个整数N,代表了硬币的种类。(1 <= N <= 500)接下来N行是N种硬币的信息,每行有两个整数P和W,分别代表价值和重量,单位是克。 输出:每例输出一行,若能装入,打印The minimum amount of money in the piggy-bank is X. 其中X是钱的总额。若不能装入,打印 ...
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. # 题意# 存钱罐里有一些钱(硬币)。所有硬币的重量已知,空存钱罐的质量e,装有钱的存钱罐的质量为f,有n行,每行代表一种硬币,每行的第一个数p表示硬币的面值,第二...
(完全背包)Piggy-Bank(P1384) 刚看了背包九讲,,做起还算轻松 #include<iostream> #include<stdlib.h> #include<math.h> #include<stdio.h> #include<algorithm> #include<queue> #include<string.h> #include<stack> #include<math.h> #include<stdlib.h>...
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weight); } int result = updateResult(fullWeight - emptyWeight, coinType); if (MAX_RESULT == result) { printf("This is impossible.\n"); } else { printf("The minimum amount of money in the piggy-bank is %d.\n", result); } } } ...
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 ...