題意：

給出一個存錢罐的重量和沒存錢之前存錢罐的重量，然后給出幾種硬幣的重量和幣值，計算存錢罐里至少有多少錢。

題目：

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.

分析：

（1）、關于恰好裝滿問題
是否恰好裝滿的解法不同只在于初始值的不同
恰好裝滿：1、求最大值時，除了dp[0] 為0，其他都初始化為無窮小 -inf;2、求最小值時，除了dp[0] 為0，其他都初始化為無窮大inf(求最小值時原狀態為inf【狀態為不存在此種情況】+一個數一定大于inf，此時選擇小值為inf，狀態不變);
不必恰好裝滿： 全初始化為0.
(2)首先滿的豬的重量減去空的豬的重量這樣就可以獲得能裝多少硬幣的重量。然后用完全背包，看是否能恰好裝滿且價值最小。
這里推薦一篇寫的好的博文寫挺好的，就是太監了的一篇博客，狗頭保命

AC代碼：

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; typedef long long ll; const int inf=0x3f3f3f3f; const int M=5e2+10; int t,a,b,n; int x[M],y[M],dp[10010]; int main() {scanf("%d",&t);while(t--){memset(dp,inf,sizeof(dp));scanf("%d%d%d",&a,&b,&n);dp[0]=0;//恰好裝滿，只有背包容量為0時滿足，其他初始化為無窮表示未被定義int ma=b-a;for(int i=1; i<=n; i++)scanf("%d%d",&x[i],&y[i]);for(int i=1; i<=n; i++)for(int j=y[i]; j<=ma; j++)//正序，放入第i物品,之前的狀態需要進行改變,故需要正序dp[j]=min(dp[j],dp[j-y[i]]+x[i]);if(dp[ma]==inf)printf("This is impossible.\n");elseprintf("The minimum amount of money in the piggy-bank is %d.\n",dp[ma]);}return 0; }

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