1、Description
Mileage program of ACM (Airline of Charming Merlion) is really nice for the travelers flying frequently. Once you complete a flight with ACM, you can earn ACMPerk miles in your ACM Mileage Bank depended on mileage you actual fly. In addition, you can use the ACMPerk mileage in your Mileage Bank to exchange free flight ticket of ACM in future. The following table helps you calculate how many ACMPerk miles you can earn when you fly on ACM. When you fly ACM Class Code You'll earn First Class F Actual mileage + 100% mileage Bonus Business Class B Actual mileage + 50% mileage Bonus Economy Class Y1-500 miles 500 miles500+ miles Actual mileageIt's shown that your ACMPerk mileage consists of two parts. One is your actual flight mileage (the minimum ACMPerk mileage for Economy Class for one flight is 500 miles), the other is the mileage bonus (its accuracy is up to 1 mile) when you fly in Business Class and First Class. For example, you can earn 1329 ACMPerk miles, 1994 ACMPerk miles and 2658 ACMPerk miles for Y, B or F class respectively for the fly from Beijing to Tokyo (the actual mileage between Beijing and Tokyo is 1329 miles). When you fly from Shanghai to Wuhan, you can earn ACMPerk 500 miles for economy class and ACMPerk 650 miles for business class (the actual mileage between Shanghai and Wuhan is 433 miles). Your task is to help ACM build a program for automatic calculation of ACMPerk mileage.
Input
The input file contains several data cases. Each case has many flight records, each per line. The flight record is in the following format: OriginalCity DistanceCity ActualMiles ClassCode Each case ends with a line of one zero. A line of one # presents the end of the input file.
Output
Output the summary of ACMPerk mileages for each test case, one per line.
Sample Input
Beijing Tokyo 1329 F
Shanghai Wuhan 433 Y
0
#
Sample Output
3158
Hint
When calculate bonus ,be sure you rounded x.5 up to x+1
2、Description
Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit even though its color and size make it indistinguishable from the real silver dollars. The counterfeit coin has a different weight from the other coins but Sally does not know if it is heavier or lighter than the real coins.
Happily, Sally has a friend who loans her a very accurate balance scale. The friend will permit Sally three weighings to find the counterfeit coin. For instance, if Sally weighs two coins against each other and the scales balance then she knows these two coins are true. Now if Sally weighs
one of the true coins against a third coin and the scales do not balance then Sally knows the third coin is counterfeit and she can tell whether it is light or heavy depending on whether the balance on which it is placed goes up or down, respectively.
By choosing her weighings carefully, Sally is able to ensure that she will find the counterfeit coin with exactly three weighings.
Input
The first line of input is an integer n (n > 0) specifying the number of cases to follow. Each case consists of three lines of input, one for each weighing. Sally has identified each of the coins with the letters A--L. Information on a weighing will be given by two strings of letters and then one of the words ``up'', ``down'', or ``even''. The first string of letters will represent the coins on the left balance; the second string, the coins on the right balance. (Sally will always place the same number of coins on the right balance as on the left balance.) The word in the third position will tell whether the right side of the balance goes up, down, or remains even.
Output
For each case, the output will identify the counterfeit coin by its letter and tell whether it is heavy or light. The solution will always be uniquely determined.
Sample Input
1
ABCD EFGH even
ABCI EFJK up
ABIJ EFGH even
Sample Output
K is the counterfeit coin and it is light.
3、Description
两只青蛙在网上相识了，它们聊得很开心，于是觉得很有必要见一面。它们很高兴地发现它们住在同一条纬度线上，于是它们约定各自朝西跳，直到碰面为止。可是它们出发之前忘记了一件很重要的事情，既没有问清楚对方的特征，也没有约定见面的具体位置。不过青蛙们都是很乐观的，它们觉得只要一直朝着某个方向跳下去，总能碰到对方的。但是除非这两只青蛙在同一时间跳到同一点上，不然是永远都不可能碰面的。为了帮助这两只乐观的青蛙，你被要求写一个程序来判断这两只青蛙是否能够碰面，会在什么时候碰面。
我们把这两只青蛙分别叫做青蛙A和青蛙B，并且规定纬度线上东经0度处为原点，由东往西为正方向，单位长度1米，这样我们就得到了一条首尾相接的数轴。设青蛙A的出发点坐标是x，青蛙B的出发点坐标是y。青蛙A一次能跳m米，青蛙B一次能跳n米，两只青蛙跳一次所花费的时间相同。纬度线总长L米。现在要你求出它们跳了几次以后才会碰面。
Input
输入只包括一行5个整数x，y，m，n，L，其中x≠y < 2000000000，0 < m、n < 2000000000，0 < L < 2100000000。
Output
输出碰面所需要的跳跃次数，如果永远不可能碰面则输出一行"Impossible"
Sample Input
1 2 3 4 5
Sample Output
4
Mileage program of ACM (Airline of Charming Merlion) is really nice for the travelers flying frequently. Once you complete a flight with ACM, you can earn ACMPerk miles in your ACM Mileage Bank depended on mileage you actual fly. In addition, you can use the ACMPerk mileage in your Mileage Bank to exchange free flight ticket of ACM in future. The following table helps you calculate how many ACMPerk miles you can earn when you fly on ACM. When you fly ACM Class Code You'll earn First Class F Actual mileage + 100% mileage Bonus Business Class B Actual mileage + 50% mileage Bonus Economy Class Y1-500 miles 500 miles500+ miles Actual mileageIt's shown that your ACMPerk mileage consists of two parts. One is your actual flight mileage (the minimum ACMPerk mileage for Economy Class for one flight is 500 miles), the other is the mileage bonus (its accuracy is up to 1 mile) when you fly in Business Class and First Class. For example, you can earn 1329 ACMPerk miles, 1994 ACMPerk miles and 2658 ACMPerk miles for Y, B or F class respectively for the fly from Beijing to Tokyo (the actual mileage between Beijing and Tokyo is 1329 miles). When you fly from Shanghai to Wuhan, you can earn ACMPerk 500 miles for economy class and ACMPerk 650 miles for business class (the actual mileage between Shanghai and Wuhan is 433 miles). Your task is to help ACM build a program for automatic calculation of ACMPerk mileage.
Input
The input file contains several data cases. Each case has many flight records, each per line. The flight record is in the following format: OriginalCity DistanceCity ActualMiles ClassCode Each case ends with a line of one zero. A line of one # presents the end of the input file.
Output
Output the summary of ACMPerk mileages for each test case, one per line.
Sample Input
Beijing Tokyo 1329 F
Shanghai Wuhan 433 Y
0
#
Sample Output
3158
Hint
When calculate bonus ,be sure you rounded x.5 up to x+1
2、Description
Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit even though its color and size make it indistinguishable from the real silver dollars. The counterfeit coin has a different weight from the other coins but Sally does not know if it is heavier or lighter than the real coins.
Happily, Sally has a friend who loans her a very accurate balance scale. The friend will permit Sally three weighings to find the counterfeit coin. For instance, if Sally weighs two coins against each other and the scales balance then she knows these two coins are true. Now if Sally weighs
one of the true coins against a third coin and the scales do not balance then Sally knows the third coin is counterfeit and she can tell whether it is light or heavy depending on whether the balance on which it is placed goes up or down, respectively.
By choosing her weighings carefully, Sally is able to ensure that she will find the counterfeit coin with exactly three weighings.
Input
The first line of input is an integer n (n > 0) specifying the number of cases to follow. Each case consists of three lines of input, one for each weighing. Sally has identified each of the coins with the letters A--L. Information on a weighing will be given by two strings of letters and then one of the words ``up'', ``down'', or ``even''. The first string of letters will represent the coins on the left balance; the second string, the coins on the right balance. (Sally will always place the same number of coins on the right balance as on the left balance.) The word in the third position will tell whether the right side of the balance goes up, down, or remains even.
Output
For each case, the output will identify the counterfeit coin by its letter and tell whether it is heavy or light. The solution will always be uniquely determined.
Sample Input
1
ABCD EFGH even
ABCI EFJK up
ABIJ EFGH even
Sample Output
K is the counterfeit coin and it is light.
3、Description
两只青蛙在网上相识了，它们聊得很开心，于是觉得很有必要见一面。它们很高兴地发现它们住在同一条纬度线上，于是它们约定各自朝西跳，直到碰面为止。可是它们出发之前忘记了一件很重要的事情，既没有问清楚对方的特征，也没有约定见面的具体位置。不过青蛙们都是很乐观的，它们觉得只要一直朝着某个方向跳下去，总能碰到对方的。但是除非这两只青蛙在同一时间跳到同一点上，不然是永远都不可能碰面的。为了帮助这两只乐观的青蛙，你被要求写一个程序来判断这两只青蛙是否能够碰面，会在什么时候碰面。
我们把这两只青蛙分别叫做青蛙A和青蛙B，并且规定纬度线上东经0度处为原点，由东往西为正方向，单位长度1米，这样我们就得到了一条首尾相接的数轴。设青蛙A的出发点坐标是x，青蛙B的出发点坐标是y。青蛙A一次能跳m米，青蛙B一次能跳n米，两只青蛙跳一次所花费的时间相同。纬度线总长L米。现在要你求出它们跳了几次以后才会碰面。
Input
输入只包括一行5个整数x，y，m，n，L，其中x≠y < 2000000000，0 < m、n < 2000000000，0 < L < 2100000000。
Output
输出碰面所需要的跳跃次数，如果永远不可能碰面则输出一行"Impossible"
Sample Input
1 2 3 4 5
Sample Output
4
if(m==n)
System.out.println("Impossible ");
else if(x<y&&m<n || x>y&&m>n)
System.out.println((x-y)/(m-n));
else if(x<y && m>n)
System.out.println(L+x-y)/(m-n);
else if(x>y && m<n)
System.out.println(L-x+y)/(n-m);