DATA SEGMENT
RESULT DB 2 DUP(?)
DATA ENDS
STACK SEGMENT
DB 200(?)
CODE SEGMENT
MAIN PROC FAR
ASSUME CS:CODE, DS:DATA
START:
PUSH DS
SUB AX,AX
PUSH AX

MOV AX,DATA
MOV DS,AX

MOV BL,0FH
AND BL,AL
PUSH BL
CALL TRANSFER
MOV AH,BL

MOV BL,11110000B
AND BL,AL
MOV CL,4
SHR BL,CL
PUSH BL
CALL TRANSFER
SHL BL,CL

MOV RESULT,AH
MOV RESULT+1,BL

MOV AH,09H
MOV DX,RESULT

INT 21H
RET
MAIN ENDP

TRANSFER:
POP BL
PUSH AX
PUSH CX
PUSH DX

CMP BL,0AH
JB PROCESS1
JNB PROCESS2
PROCESS1:
PUSH BL
JMP QUIT
PROCESS2:
POP AX
POP CX
POP DX
PUSH BL
QUIT:
RET
END TRANSFER

CODE EDNS
END START

fatal error : attemp to access data outside the segment

1个回答

Think I’ll Buy Me a Football Team 状态切换问题
Problem Description Falling Stocks. Bankrupted companies. Banks with no Cash. Seems like the best time to invest: ``Think I'll buy me a football team!" No seriously, I think I have the solution to at least the problem of cash in banks. Banks nowadays are all owing each other great amounts of money and no bank has enough cash to pay other banks' debts even though, on paper at least, they should have enough money to do so. Take for example the inter-bank loans shown in figure (a). The graph shows the amounts owed between four banks (A ...D). For example, A owes B 50M while, at the same time, B owes A 150M. (It is quite common for two banks to owe each other at the same time.) A total amount of 380M in cash is needed to settle all debts between the banks. In an attempt to decrease the need for cash, and after studying the example carefully, I concluded that there's a lot of cash being transferred unnecessarily. Take a look: 1. C owes D the same amount as D owes A, so we can say that C owes A an amount of 30M and get D out of the picture. 2. But since A already owes C 100M, we can say that A owes C an amount of 70M. 3. Similarly, B owes A 100M only, (since A already owes B 50M.) This reduces the above graph to the one shown in figure (b) which reduces the needed cash amount to 190M (A reduction of 200M, or 53%.) 4. I can still do better. Rather than B paying A 100M and A paying 70M to C, B can pay 70M (out of A's 100M) directly to C. This reduces the graph to the one shown in figure (c). Banks can settle all their debts with only 120M in cash. A total reduction of 260M or 68%. Amazing! I have data about inter-bank debts but I can't seem to be able to process it to obtain the minimum amount of cash needed to settle all the debts. Could you please write a program to do that? Input Your program will be tested on one or more test cases. Each test case is specified on N + 1 lines where N < 1, 000 is the number of banks and is specified on the first line. The remaining N lines specifies the inter-bank debts using an N×N adjacency matrix (with zero diagonal) specified in row-major order. The ith row specifies the amounts owed by the ith bank. Amounts are separated by one or more spaces. All amounts are less than 1000. The last line of the input file has a single 0. Output For each test case, print the result using the following format: k . B A where k is the test case number (starting at 1,) is a space character, B is the amount of cash needed before reduction and A is the amount of cash after reduction. Sample Input 4 0 50 100 0 150 0 20 0 0 0 0 30 30 0 0 0 0 Sample Output 1. 380 120
Musical Chairs 音乐链条的问题
Problem Description In the traditional game of Musical Chairs, N + 1 children run around N chairs (placed in a circle) as long as music is playing. The moment the music stops, children run and try to sit on an available chair. The child still standing leaves the game, a chair is removed, and the game continues with N children. The last child to sit is the winner. In an attempt to create a similar game on these days' game consoles, you modify the game in the following manner: N Children are seated on N chairs arranged around a circle. The chairs are numbered from 1 to N . Your program pre-selects a positive number D . The program starts going in circles counting the children starting with the first chair. Once the count reaches D , that child leaves the game, removing his/her chair. The program starts counting again, beginning with the next chair in the circle. The last child remaining in the circle is the winner. For example, consider the game illustrated in the figure above for N = 5 and D = 3 . In the figure, the dot indicates where counting starts and × indicates the child leaving. Starting off, child #3 leaves the game, and counting restarts with child #4. Child #1 is the second child to leave and counting restart with child #2 resulting in child #5 leaving. Child #2 is the last to leave, and child #4 is the winner. Write a program to determine the winning child given both N and D . Input Your program will be tested on one or more test cases. Each test case specifies two positive integers N and D on a single line, separated by one or more spaces, where N, D < 1, 000, 000 . The last line of the input file contains two 0's and is not part of the test cases. Output For each test case, write the winner using the following format: N D W Where N and D are as above, is a space character, and W is the winner of that game. Sample Input 5 3 7 4 0 0 Sample Output 5 3 4 7 4 2
matlab拿别人的程序跑时出现Attempt to reference field of non-structure array.

Crash and Go(relians) 代码的写法
Think I’ll Buy Me a Football Team 球队问题
Problem Description Falling Stocks. Bankrupted companies. Banks with no Cash. Seems like the best time to invest: ``Think I'll buy me a football team!" No seriously, I think I have the solution to at least the problem of cash in banks. Banks nowadays are all owing each other great amounts of money and no bank has enough cash to pay other banks' debts even though, on paper at least, they should have enough money to do so. Take for example the inter-bank loans shown in figure (a). The graph shows the amounts owed between four banks (A ...D). For example, A owes B 50M while, at the same time, B owes A 150M. (It is quite common for two banks to owe each other at the same time.) A total amount of 380M in cash is needed to settle all debts between the banks. In an attempt to decrease the need for cash, and after studying the example carefully, I concluded that there's a lot of cash being transferred unnecessarily. Take a look: 1. C owes D the same amount as D owes A, so we can say that C owes A an amount of 30M and get D out of the picture. 2. But since A already owes C 100M, we can say that A owes C an amount of 70M. 3. Similarly, B owes A 100M only, (since A already owes B 50M.) This reduces the above graph to the one shown in figure (b) which reduces the needed cash amount to 190M (A reduction of 200M, or 53%.) 4. I can still do better. Rather than B paying A 100M and A paying 70M to C, B can pay 70M (out of A's 100M) directly to C. This reduces the graph to the one shown in figure (c). Banks can settle all their debts with only 120M in cash. A total reduction of 260M or 68%. Amazing! I have data about inter-bank debts but I can't seem to be able to process it to obtain the minimum amount of cash needed to settle all the debts. Could you please write a program to do that? Input Your program will be tested on one or more test cases. Each test case is specified on N + 1 lines where N < 1, 000 is the number of banks and is specified on the first line. The remaining N lines specifies the inter-bank debts using an N×N adjacency matrix (with zero diagonal) specified in row-major order. The ith row specifies the amounts owed by the ith bank. Amounts are separated by one or more spaces. All amounts are less than 1000. The last line of the input file has a single 0. Output For each test case, print the result using the following format: k . B A where k is the test case number (starting at 1,) is a space character, B is the amount of cash needed before reduction and A is the amount of cash after reduction. Sample Input 4 0 50 100 0 150 0 20 0 0 0 0 30 30 0 0 0 0 Sample Output 1. 380 120
Cattle 的程序的编写
Snooker Referee 裁判的问题

Snooker Referee 比赛裁判的算法
Crash and Go(relians) 怎么实现的
The Embarrassed Cryptographer 的程序的设计
Problem Description The young and very promising cryptographer Odd Even has implemented the security module of a large system with thousands of users, which is now in use in his company. The cryptographic keys are created from the product of two primes, and are believed to be secure because there is no known method for factoring such a product effectively. What Odd Even did not think of, was that both factors in a key should be large, not just their product. It is now possible that some of the users of the system have weak keys. In a desperate attempt not to be fired, Odd Even secretly goes through all the users keys, to check if they are strong enough. He uses his very poweful Atari, and is especially careful when checking his boss' key. Input The input consists of no more than 20 test cases. Each test case is a line with the integers 4 <= K <= 10100 and 2 <= L <= 106. K is the key itself, a product of two primes. L is the wanted minimum size of the factors in the key. The input set is terminated by a case where K = 0 and L = 0. Output For each number K, if one of its factors are strictly less than the required L, your program should output "BAD p", where p is the smallest factor in K. Otherwise, it should output "GOOD". Cases should be separated by a line-break. Sample Input 143 10 143 20 667 20 667 30 2573 30 2573 40 0 0 Sample Output GOOD BAD 11 GOOD BAD 23 GOOD BAD 31
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