Piggy-Bank (完全背包)
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
Output
Sample Input
Sample Output
The minimum amount of money in the piggy-bank is 100.
This is impossible.
#include <stdio.h>
#define INF 0x3f3f3f3f
#include <algorithm>
using namespace std; int t, w[], val[], dp[], w0, w1; //w0存空存钱罐重量,w1存装满的存钱罐重量,dp[i]代表容量为i的时候所能存的最少钱数,t钱币种类数
void work()
{
for(int i = ; i <= w1 - w0; i++)
dp[i] = INF; // 初始标记为无穷,如果计算完之后任然为无穷的话说明无解,小于无穷则输出结果
dp[] = ;//容量为0所能存的资金也是0
for(int i = ; i < t; i++)
{
for(int j = w[i]; j <= w1 - w0; j++)
dp[j] = min(dp[j], dp[j-w[i]] + val[i]);//算出只放i种钱币,其中第i种钱币放(0个——所能放的最多数量)的时候存钱罐里所存的的最少资金
}
}
int main()
{ int n;
scanf("%d", &n);
while(n--)
{
scanf("%d%d", &w0, &w1);
scanf("%d", &t);
for(int i = ; i < t; i++)
{
scanf("%d%d", &val[i], &w[i]);
}
work();
if(dp[w1 - w0] < INF)
printf("The minimum amount of money in the piggy-bank is %d.\n", dp[w1 - w0]);
else
printf("This is impossible.\n");
}
return ;
}
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