POJ 2115 C Looooops【Exgcd+位移（LL）】

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本文介绍了一种计算特定C语言循环执行次数的方法，该循环在每次迭代时都会对变量进行模2^k的操作。文章提供了详细的算法思路及AC代码实现，并强调了在处理大数值时应注意的数据类型选择。

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F - C Looooops

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

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 < 2 k) modulo 2 k.

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 < 2 k) 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

题意: 求上述循环运行多少次（每次运算要取模）；

思路：如果不取模就是B=A mod (C) 的整数解，取模就是每次加C的时候对2^K取模，B=(A+Cx) mod (2^K) 求出x的最小非负解就是答案；

失误：思路没有问题，但是实现出了问题：2^K=1<<K 但是这是int的时候，这个题恰好K<=32 int是有符号的1<<32=0,这时候就不对了，用LL : 2^K=1LL<<K; __int64 : 2^K=(__int64 )1<<K;

AC代码：

#include<cstdio>

typedef long long LL;

LL Exgcd(LL a,LL b,LL &x,LL &y)
{
	if(!b)
	{
		x=1, y=0;
		return a;
	 } 
	 LL r=Exgcd(b,a%b,y,x);
	 y-=a/b*x;
	 return r;
}

int main()
{
	LL A,B,C,K;
	while(~scanf("%lld %lld %lld %lld",&A,&B,&C,&K))
	{
		if(!A&&!B&&!C&&!K)  break;
		LL x=0,y=0;
		LL d=Exgcd(C,1LL<<K,x,y);
		if((B-A)%d) {
			printf("FOREVER\n");
			continue;
		} 
		x=(B-A)/d*x;
		LL mod=(1LL<<K)/d;
		x=(x%mod+mod)%mod;
		printf("%lld\n",x);
	}
	return 0;
 }