[2020.10.26 SSL Simulation Tournament T1] Magic Tribe [Number Theory Inverse Element]

table of Contents:

T1：魔法部落
T2：圆盘
T3：棋盘行走
T4：走方格

A set of questions that are not difficult as a whole but require details

T1 ：

analysis:

It can be seen that this is a common ratio of 3 3The geometric sequence of$3$ is
known to be the firstnnof the geometric sequence$The\; sum\; of\; n$ items is: common ratio$q$^$(n+1)/($ Common ratio$q-1)$
That is:$3$^$(n+1)/2$
So this quick power$+mod\; can$ welcome$60pts$ and$wa$ some point
why?
Because$x/y$再$mod$ may appear negative, so it needs to be solved by inverse element.
Finally:$q$^$(n+1)/(q-1)mod(1e9+7)$
$=q$ ^$(n+1)\ast (1e9+7+1)/2$ mod$(1e9+7)$
$=q$^$(n+1)\ast 5e8+4$ mod$(1e9+7)$

CODE：

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const long long p=1e9+7;
const long long s=5e8+4;
ll n,ans,qwq=1;
void ksm(ll n)
{
    
    
	long long tmp=3;
    while(n){
    
    
        if(n&1){
    
    
            qwq=(qwq*tmp)%p;
        }
        tmp=(tmp*tmp)%p;
        n>>=1;
    }
}
int main(){
    
    
	scanf("%lld",&n);
	ksm(n+1);
	ans=(qwq-1)*s%p;
	printf("%lld",ans);
	return 0;
}