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!

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!

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.

the weight cannot be reached exactly, print a line “This is impossible.”.

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

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.

简单的装满背包的最小价值。

杭电机制，数组开小了会TLE QAQ。

#includeusing namespace std; const int MAXN = 1e3+7; const int inf = 1e9; int n,m; int p[MAXN],w[MAXN]; int dp[MAXN*10]; int e,f; int main() { int t; scanf("%d",&t); while(t--) { scanf("%d%d",&e,&f); f -= e; for(int i = 0 ; i <= f ; ++i)dp[i] = -inf; scanf("%d",&n); for(int i = 0 ; i < n ; ++i) { scanf("%d%d",&p[i],&w[i]); p[i] = -p[i]; } dp[0] = 0; for(int i = 0 ; i < n ; ++i) for(int j = w[i] ; j <= f ; ++j) { dp[j] = max(dp[j],dp[j-w[i]] + p[i]); } if(dp[f] == -inf)puts("This is impossible."); else printf("The minimum amount of money in the piggy-bank is %d.\n",-dp[f]); } return 0; }

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