hdu4565 So Easy! 矩阵快速幂

A sequence S_n is defined as:

Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate S_n.
　　You, a top coder, say: So easy!

矩阵快速幂

 #include<stdio.h>
 #include<string.h>
 #include<math.h>
 typedef long long ll;
 int mod;

 struct mat{
     int r,c;
     ll m[][];        //经测试最大开成590*590的 ll 型矩阵
     mat(){}
     mat(int r,int c):r(r),c(c){}
     void clear(){
         memset(m,,sizeof(m));
     }

     mat operator+(mat a)const{
         mat ans(r,c);
         for(int i=;i<=r;i++){
             for(int j=;j<=c;j++){
                 ans.m[i][j]=(m[i][j]+a.m[i][j])%mod;
             }
         }
         return ans;
     }

     mat operator*(mat a)const{
         mat tmp(r,a.c);
         int i,j,k;
         for(i=;i<=tmp.r;i++){
             for(j=;j<=tmp.c;j++){
                 tmp.m[i][j]=;
                 for(k=;k<=c;k++){
                     tmp.m[i][j]=(tmp.m[i][j]+(m[i][k]*a.m[k][j])%mod)%mod;
                 }
             }
         }
         return tmp;
     }

     mat operator^(int n)const{        //需要时可以用 ll n，注意运算符优先级比较低，多用括号；
         mat ans(r,r),tmp(r,r);
         memcpy(tmp.m,m,sizeof(tmp.m));
         ans.clear();
         for(int i=;i<=ans.r;i++){
             ans.m[i][i]=;
         }
         while(n){
             if(n&)ans=ans*tmp;
             n>>=;
             tmp=tmp*tmp;
         }
         return ans;
     }

     void print()const{
         for(int i=;i<=r;i++){
             for(int j=;j<=c;j++){
                 printf("%lld",m[i][j]);
                 if(j==c)printf("\n");
                 else printf(" ");
             }
         }
     }

 };

 int main(){
     ll a,b,n;
     while(scanf("%lld%lld%lld%d",&a,&b,&n,&mod)!=EOF){
         if(n==){
             printf("%lld\n",(*a)%mod);
             continue;
         }
         mat tmp(,),t(,);
         tmp.m[][]=(*a)%mod;
         tmp.m[][]=((b%mod-(a*a)%mod)%mod+mod)%mod;
         tmp.m[][]=;
         tmp.m[][]=;
         t.m[][]=((*a)%mod)*a%mod+(*(b%mod))%mod;
         t.m[][]=(*a)%mod;
         t=(tmp^(n-))*t;
         printf("%lld\n",t.m[][]);
     }
     return ;
 }