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!!!!!!!!!!!!!! "); 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 ",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 ",mi ); 113 //continue; 114 printf("%lld ",get_ans(mi)); 115 } 116 return 0; 117 }