HDU 2604
快速幂取模,快速矩阵幂取模
先利用DFA构造转移矩阵
[cpp]
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
int len,mod;
ll tem[6][6];
int main(){
int i,j,k;
while(scanf("%d %d",&len,&mod)==2){
ll a[6][6]={{0,0,1,0,0,1},{0,0,0,0,0,1},{0,0,0,0,1,0},{0,0,0,1,1,0},{0,2,0,1,0,0},{2,0,1,0,0,0}};
ll t[6][6]={{1,0,0,0,0,0},{0,1,0,0,0,0},{0,0,1,0,0,0},{0,0,0,1,0,0},{0,0,0,0,1,0},{0,0,0,0,0,1}};
int l=len;
while(l){
if(l%2){
for(i=0;i<=5;i++)
for(j=0;j<=5;j++)
for(k=0;k<=5;k++)
tem[i][j]+=(t[i][k]*a[k][j])%mod;
for(i=0;i<=5;i++)
for(j=0;j<=5;j++){
t[i][j]=tem[i][j]%mod;
tem[i][j]=0;
}
}
for(i=0;i<=5;i++)
for(j=0;j<=5;j++)
for(k=0;k<=5;k++)
tem[i][j]+=(a[i][k]*a[k][j])%mod;
for(i=0;i<=5;i++)
for(j=0;j<=5;j++){
a[i][j]=tem[i][j]%mod;
tem[i][j]=0;
}
l/=2;
}
ll ans=1;
ll m=2;
while(len){
if(len%2)
ans=(ans*m)%mod;
m=(m*m)%mod;
len/=2;
}
ans%=mod;
printf("%I64d\n",(ans+mod*2-t[5][5]-t[4][5])%mod);
}
return 0;
}
补充:软件开发 , C++ ,