C looooops (POJ 2115)

Approximate test instructions:

For C's for (i=a; i!=b; I +=c) loop, ask how many loops in the K-bit storage system will end.

If the end is within a finite number of times, the output cycle times.

Otherwise the output is dead loop.

Problem Solving Ideas:

Test instructions is not difficult to understand, but the use of K-bit storage system data characteristics to cycle.

For example , the int is 16 bits, then int can save 2^16 data, that is, the maximum number is 65535 (the default is unsigned),

When the loop causes I to exceed 65535, I will return 0 to start counting again

such as i=65534, when i+=3,i=1

is actually i= (65534+3)% (2^16) =1

With these ideas, it is easy to get the equation for a set of data to be looped through X-times:

x=[(b-a+2^k)%2^k]/C

That is cx= (b-a) (mod 2^k) This equation is a modal equation, the subject is to find the value of x.

//note the formula that finally changes the x to the smallest positive integer#include <cstdio>#include<iostream>#definell Long Longusing namespacestd;ll a,b,c,k;ll e_gcd (ll a,ll b,ll&x,ll &y) {    if(b==0) {x=1; y=0; returnA; } ll R=E_GCD (b,a%b,x,y); ll T=x;x=y;y=t-a/b*y; returnR;}intMain () { while(1) {cin>>A>>B>>C>>J; if(a==0&&b==0&&c==0&&k==0) Break; if(a==B) {printf ("0\n"); Continue; } ll a=c,b=1ll<<k,c=b-A,x,y; ll GCD=E_GCD (a,b,x,y); if(c%gcd) {printf ("forever\n"); Continue; } x=x*c/GCD; X= (x (B/GCD) + (B/GCD))% (b/GCD); cout<<x<<Endl; }}

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C looooops (POJ 2115)