B. Jzzhu and Sequences

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=100000+10;
const ll MOD =1000000007 ;
struct Mar
{
    ll mat[2][2];
    Mar operator *(const Mar &m )const
    {
        Mar tmp;
        for(int i=0;i<2;i++)
        {
            for(int j=0;j<2;j++)
            {
                tmp.mat[i][j]=0;
                for(int k=0;k<2;k++)
                {
                    tmp.mat[i][j]=(tmp.mat[i][j]%MOD+mat[i][k]*m.mat[k][j]%MOD)%MOD;
                    tmp.mat[i][j]%=MOD;
                }
            }
        }
        return tmp;
    }
};
Mar quick(Mar a,ll b)
{
    Mar ans;
    memset(ans.mat,0,sizeof(ans.mat));
    for(int i=0;i<2;i++)ans.mat[i][i]=1;
    while(b)
    {
        if(b&1)
            ans=ans*a;
        b>>=1;
        a=a*a;
    }
    return ans;
}
int main()
{
    ll x,y,n;
    scanf("%lld%lld%lld",&x,&y,&n);
    Mar a;
    memset(a.mat,0,sizeof(a.mat));
    if(n==1)
    {
        printf("%lld\n",(x+MOD)%MOD);
        return 0;
    }
    a.mat[0][0]=1;
    a.mat[1][0]=1;
    a.mat[0][1]=-1;
    a.mat[1][1]=0;
    Mar ans=quick(a,n-2);
   // printf("%lld %lld\n",ans.mat[1][0],ans.mat[1][1]);
    printf("%lld\n",((y%MOD*ans.mat[0][0]%MOD+MOD)%MOD+(x%MOD*ans.mat[0][1]%MOD+MOD)%MOD+MOD)%MOD);
    return 0;
}

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转载自blog.csdn.net/pandauncle/article/details/80209812