2020 Niu Ke NOIP pre-match training camp-increase the number of girls in the T1 Niu Banxian group (the third game)

Niu Banxian has three sisters, their attribute values ​​are $A$,$B$,$C$ . These three girls are telepathic, and when one person's attribute value changes, the other person's attribute value also changes.
The law of change is as follows: Suppose the attribute value of the first girl is$A$ , the attribute value of the second girl is$B$ , the attribute value of the third girl is$C$，

• If $A+B\le C$ then the attribute value of the first girl becomes$2A$ , the attribute value of the second girl becomes$2B$ , the attribute value of the third girl becomes$C-A-B$。
• Otherwise set $A$,$The$ smaller attribute value in$B$ is$W$，$P=min(\lfloor C/2\rfloor ,W-1)$ . Then the attribute value of the first girl becomes$A-P$ , the attribute value of the second girl becomes$B+P-C$ , the attribute value of the third girl becomes$2C$ .
  Niu Banxian wants to know that the third girl passed byKKThe attribute value after$K$ changes. Because Niu Banxian wants to talk about life with his sister, he left this question to you.
  Link: https://ac.nowcoder.com/acm/contest/7609/A
  Source: Niuke

Simplify decently given two numbers $a(a=A+B+C)$ and$b(b=C)$
若 $b>a$ 则 $b=b-a,a=2\cdot a$
否则 $b=2\cdot b,a=a-b$

Discussing the value classification of b can conclude that
$ans=(2^k\cdot b)mod(a+b)$

#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
ull t,u,v,mod,b,k,ans;
inline ull ksm(ull x,ull y){
    
    
	ull res=1;
	while(y){
    
    
		if(y&1)res=res*x%mod;
		x=x*x%mod; y>>=1;
	} 
	return res;
}
int main(){
    
    
	scanf("%llu",&t);
	while(t--){
    
    
		scanf("%llu %llu %llu %llu",&u,&v,&b,&k);
		mod=u+v+b;
		ans=ksm(2,k)*b%mod;
		printf("%llu\n",ans);
	}
}