HDU 5895 Mathematician QSC

题目地址

欧拉函数+矩阵快速幂

  1 #include<cstdio>
  2 #include<algorithm>
  3 #include<string.h>
  4 #include<queue>
  5 #define LL long long
  6 using namespace std;
  7 const int Nmax=10;
  8 LL n,y,x,s,tmp;
  9 int mod;
 10 int oula_mod;
 11 
 12 struct Matrix
 13 {
 14     int n,m;
 15     long long map[Nmax][Nmax];
 16     Matrix(int x,int y)
 17     {
 18         n=x;m=y;
 19         for(int i=1;i<=n;i++)
 20             for(int j=1;j<=m;j++)
 21                 map[i][j]=0;
 22     }
 23     Matrix operator * (const Matrix b)
 24     {
 25         Matrix c(n,b.m);
 26         if(m==b.n)
 27         {
 28             for(int i=1;i<=c.n;i++)
 29                 for(int j=1;j<=c.m;j++)
 30                     for(int k=1;k<=m;k++)
 31                         c.map[i][j]=(c.map[i][j]+(map[i][k]*b.map[k][j])%oula_mod)%oula_mod;
 32             return c;
 33         }
 34         printf("error!!!!!!!!!!!!!!\n");    
 35     }
 36 };
 37 
 38 
 39 int oula(int n)
 40 {
 41     int ret=1,i;
 42     for(i=2;i*i<=n;i++)
 43     {
 44         if(n%i==0)
 45         {
 46             n/=i,ret*=i-1;
 47             while(n%i==0) n/=i,ret*=i;
 48         }
 49     }
 50     if(n>1) ret*=n-1;
 51     return ret;
 52 }
 53 
 54 
 55 
 56 
 57 Matrix get(long long n)
 58 {
 59     Matrix base(4,4);
 60     base.map[1][1]=1;base.map[1][2]=1;base.map[1][3]=0;base.map[1][4]=0;
 61     base.map[2][1]=0;base.map[2][2]=4;base.map[2][3]=1;base.map[2][4]=4;
 62     base.map[3][1]=0;base.map[3][2]=1;base.map[3][3]=0;base.map[3][4]=0;
 63     base.map[4][1]=0;base.map[4][2]=2;base.map[4][3]=0;base.map[4][4]=1;
 64     Matrix ans(4,4);
 65     for(int i=1;i<=ans.n;i++)
 66         ans.map[i][i]=1;
 67     
 68     while(n>0)
 69     {
 70         if(n & 1)
 71             ans=ans*base;
 72         base=base*base;
 73         n>>=1;
 74     }
 75     
 76     return ans;
 77 }
 78 
 79 long long get_ans(long long times)
 80 {
 81     long long ans=1;
 82     long long base=x;
 83     while(times>0)
 84     {
 85         if(times & 1)
 86             ans=(ans*base)%mod;
 87         base=(base*base)%mod;
 88         times>>=1;
 89     }
 90     return ans;
 91 }
 92 
 93 
 94 int main()
 95 {
 96     
 97     int t;
 98     scanf("%d",&t);
 99     while(t--)
100     {
101         scanf("%lld%lld%lld%lld",&n,&y,&x,&s);
102         mod=s+1;
103         oula_mod=oula(mod);
104         //printf("oula_mod:%d\n",oula_mod);
105         Matrix base(4,1);
106         base.map[1][1]=0;
107         base.map[2][1]=1;
108         base.map[3][1]=0;
109         base.map[4][1]=0;
110         Matrix ans=get(n*y)*base;
111         long long mi=ans.map[1][1]+oula_mod;
112         //printf("mi:%lld\n",mi );
113         //continue;
114         printf("%lld\n",get_ans(mi));
115     }
116     return 0;
117 }

 

posted @ 2016-10-07 01:28  BBBob  阅读(158)  评论(0)  编辑  收藏  举报