标签：poj
C Looooops
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 19141 |
|
Accepted: 5049 |
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
< 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
设步数为x，原式可以变成：A+C*x==B (mod M) -->C*x+M*y=B-A (M为2的k次幂)
进一步变换成:a*x + b*y =c. (a=C,b=M,c=B-A),
解线性同余方程，方程无解则是FOREVER
#include <stdio.h>
#include <string.h>
#include <math.h>
typedef __int64 ll;
void Exgcd(ll a,ll b,ll &d,ll &x,ll &y){
if(b==0){
x=1;y=0;d=a;return ;
}
Exgcd(b,a%b,d,y,x);
y-=(a/b)*x;
}
int main()
{
ll a,b,c,d,g,k,x,y;
while(scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&k)!=EOF){
if(a+b+c+k==0) break;
ll temp=c;
c=b-a;
a=temp;
b=(ll)pow(2,k);
Exgcd(a,b,g,x,y);
if(c%g){
printf("FOREVER\n");
continue;
}
x=((x*c/g)%(b/g)+(b/g))%(b/g);
printf("%I64d\n",x);
}
return 0;
}
POJ 2115 C Looooops (线性同余方程)
标签：poj
原文地址：http://blog.csdn.net/u013068502/article/details/45176825
