题意：
经过一番推导之后，简单来说就是求2^n - 2的值，要用到快速乘法。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<cstring>
using namespace std;
#define ll long long
ll powmul(ll a, ll b, ll c){
ll ans = 0;
a = a % c;
while(b > 0){
if(b % 2) ans = (ans + a)% c;
b = b / 2;
a = (a * 2) % c;
}
return ans;
}
ll powmod(ll a,ll b,ll c){
ll ans=1;
a = a % c;
while(b>0){
if(b%2) ans = powmul(ans, a, c);//不用快速乘法的话，ll会数据溢出
b = b/2;
a = powmul(a, a, c);//同上
}
return ans;
}
int main(){
ll n,p;
while(scanf("%I64d%I64d",&n,&p)!=EOF){
ll te = 2 % p;
if(n == 1) printf("%I64d\n",n%p);
else printf("%I64d\n",(powmod(2,n,p)+p-te) % p);//注意不能取余之后相减
}
return 0;
}
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