推出公式，然后用矩阵快速幂做一下就好了。。。
#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 10005
#define maxm 40005
#define eps 1e-10
#define mod 1000000007
#define INF 999999999
#define lowbit(x) (x&(-x))
#define mp mark_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
//typedef vector<int>::iterator IT;
typedef long long LL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
// head
//POJ 1330
LL mat[5][5], mid[5][5];
LL res[5][5];
LL t, n;
LL m;
void work(void)
{
mat[1][1] = 4;
mat[1][2] = 4;
mat[2][1] = 0;
mat[2][2] = 1;
memset(res, 0, sizeof res);
for(int i = 1; i <= 2; i++) res[i][i] = 1;
while(t) {
if(t%2) {
for(int i = 1; i <= 2; i++)
for(int j = 1; j <= 2; j++) {
LL tmp = 0;
for(int k = 1; k <= 2; k++) tmp = (tmp + mat[i][k] * res[k][j]) % m;
mid[i][j] = tmp;
}
for(int i = 1; i <= 2; i++)
for(int j = 1; j <= 2; j++)
res[i][j] = mid[i][j];
}
t /= 2;
for(int i = 1; i <= 2; i++)
for(int j = 1; j <= 2; j++) {
LL tmp = 0;
for(int k = 1; k <= 2; k++) tmp = (tmp + mat[i][k] * mat[k][j]) % m;
mid[i][j] = tmp;
}
for(int i = 1; i <= 2; i++)
for(int j = 1; j <= 2; j++)
mat[i][j] = mid[i][j];
}
}
int main(void)
{
while(scanf("%I64d%I64d", &n, &m)!=EOF) {
if(n%2 == 1) t = n/2;
else t = (n-1)/2;
work();
LL ans = (res[1][2] + 1) % m;
if(n%2 == 0) ans = (ans * 2) % m;
printf("%I64d\n", ans);
}
return 0;
}
