hdu4990 Reading comprehension 矩阵快速幂

Read the program below carefully then answer the question.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include<iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include<vector>

const int MAX=100000*2;
const int INF=1e9;

int main()
{
  int n,m,ans,i;
  while(scanf("%d%d",&n,&m)!=EOF)
  {
    ans=0;
    for(i=1;i<=n;i++)
    {
      if(i&1)ans=(ans*2+1)%m;
      else ans=ans*2%m;
    }
    printf("%d\n",ans);
  }
  return 0;
}

矩阵快速幂

 #include<stdio.h>
 #include<string.h>
 typedef long long ll;
 int mod;
 struct mat{
     int r,c;
     ll m[][];
     void clear(){
         for(int i=;i<=r;i++)memset(m[i],,sizeof(m[i]));
     }
 };

 mat MatMul(mat m1,mat m2){
     mat tmp;
     tmp.r=m1.r;
     tmp.c=m2.c;
     int i,j,k;
     for(i=;i<=tmp.r;i++){
         for(j=;j<=tmp.c;j++){
             ll t=;
             for(k=;k<=m1.c;k++){
                 t=(t+(m1.m[i][k]*m2.m[k][j])%mod)%mod;
             }
             tmp.m[i][j]=t;
         }
     }
     return tmp;
 }

 mat MatQP(mat a,int n){
     mat ans,tmp=a;
     ans.r=ans.c=a.r;
     memset(ans.m,,sizeof(ans.m));
     for(int i=;i<=ans.r;i++){
         ans.m[i][i]=;
     }
     while(n){
         if(n&)ans=MatMul(ans,tmp);
         n>>=;
         tmp=MatMul(tmp,tmp);
     }
     return ans;
 }

 int main(){
     mat a;
     a.r=;a.c=;
     a.m[][]=;
     a.m[][]=;
     a.m[][]=;
     mat t;
     t.r=;
     t.c=;
     t.clear();
     t.m[][]=;
     t.m[][]=;
     t.m[][]=;
     t.m[][]=;
     t.m[][]=;
     int n;
     while(scanf("%d%d",&n,&mod)!=EOF){
         mat tmp=MatQP(t,n/);
         tmp=MatMul(tmp,a);
         printf("%lld\n",tmp.m[n%+][]);
     }
     return ;
 }