[原创]POJ 2115 C Looooops [扩展欧几里得]【数论】[水]

2016-09-13 12:29:47 Tabris_ 阅读数：405

https://blog.csdn.net/qq_33184171/article/details/52524819

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C Looooops
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 24028 Accepted: 6658
Description

A Compiler Mystery: We are given a C-language style for loop of type
for (variable = A; variable != B; variable += C)

statement;

I.e., a loop which starts by setting variable to value A and while variable is not equal to B, repeats statement followed by increasing the variable by C. We want to know how many times does the statement get executed for particular values of A, B and C, assuming that all arithmetics is calculated in a k-bit unsigned integer type (with values 0 <= x < 2k) modulo 2k.

Input

The input consists of several instances. Each instance is described by a single line with four integers A, B, C, k separated by a single space. The integer k (1 <= k <= 32) is the number of bits of the control variable of the loop and A, B, C (0 <= A, B, C < 2k) are the parameters of the loop.

The input is finished by a line containing four zeros.
Output

The output consists of several lines corresponding to the instances on the input. The i-th line contains either the number of executions of the statement in the i-th instance (a single integer number) or the word FOREVER if the loop does not terminate.
Sample Input

3 3 2 16
3 7 2 16
7 3 2 16
3 4 2 16
0 0 0 0
Sample Output

0
2
32766
FOREVER
Source

CTU Open 2004

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for (variable = A; variable != B; variable += C)

a+xc=b+y(2^k)

b-a=xc-y(2^k)

(c,2^k)=xc-y(2^k)

(b-a)/d
xc-(b-a)/ry*(2^k)=b-a

x = (b-a)/d*x;

x = (x%(2^k)+(2^k))%(2^k);

if(x==0) x=2^k;

!!!最最重要的是POJ交G++要用I64 然而VJ上告诉的是lld WA了无数发 还是最后看了代码才知道5555555555555

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