hdoj 1588 矩阵快速幂 二分等比数列求和模板

等比数列求和可以二分

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define LL long long
using namespace std;
LL mod,n,k,b;
//快速幂
struct ma{
     LL m[4][4],row,col;
     ma()
     {
         memset(m,0,sizeof(m));
         row = col = 0;
     }
     ma operator * (ma ma1) //乘
     {
        ma ans;
        ans.row = row;
        ans.col = ma1.col;
        for(int i = 1;i <= ans.row;i++)
            for(int j = 1;j <= ans.col;j++)
                for(int k = 1;k <= col;k++)
                {
                    ans.m[i][j] += (m[i][k] * ma1.m[k][j]);
                    ans.m[i][j] %= mod;
                }
        return ans;
     }
     ma operator + (ma ma1) //加
     {
         ma ans;
         ans.row = row;
         ans.col = col;
         for(int i = 1;i <= row;i++)
             for(int j = 1;j <= col;j++)
                 {
                    ans.m[i][j] =  ma1.m[i][j] + m[i][j];
                    ans.m[i][j] %= mod;
                 }
          return  ans;
     }
     ma operator^(LL n)  // 快速幂
     {
         ma ans,ma1;
         ans.row = row; ans.col = col;
         ans.m[1][1] = 1;ans.m[1][2] = 0;
         ans.m[2][1] = 0;ans.m[2][2] = 1;

       //  printf("%d\n",n);
        // ans.debug();

         ma1.row = row;  ma1.col = col;
         ma1.m[1][1] = m[1][1];ma1.m[1][2] = m[1][2];
         ma1.m[2][1] = m[2][1];ma1.m[2][2] = m[2][2];

         while(n)
         {

            if(n&1)ans = ans * ma1;
            ma1 = ma1 * ma1;
            n >>= 1;
         }
         return ans;
     }
     void debug()
     {
         printf("asda %I64d %I64d\n",row,col);
         for(int i = 1;i <= row;i++)
         {
             for(int j = 1;j <= col;j++)
                 printf("%I64d ",m[i][j]);
             printf("\n");
         }
     }
}E,mul;
ma bin(ma ma1,LL n) //求 ma^1 到 ma^n 次方的和没有ma^0 主函数还要加个1 
{
   if(n==1)return ma1;
   if(n&1)return (ma1^n) + bin(ma1,n-1); //
   else return bin(ma1,n/2) * ((ma1^(n/2)) + E);
}
int main()
{
    E.row = 2;E.col = 2;
    E.m[1][1] = 1;E.m[1][2] = 0;
    E.m[2][1] = 0;E.m[2][2] = 1;

    mul.row = 2;mul.col = 2;
    mul.m[1][1] = 1;mul.m[1][2] = 1;
    mul.m[2][1] = 1;mul.m[2][2] = 0;

    while(scanf("%I64d%I64d%I64d%I64d",&k,&b,&n,&mod)!=EOF)
    {
          ma ma1 = (mul ^ (b));

          ma ma2 = (mul ^ (k));

          if(n-1>=1)ma2 = bin(ma2,n-1);
          ma2 = ma2 + E;
          ma ans;
          ans.row = 1; ans.col = 2;
          ans.m[1][1] = 0, ans.m[1][2] = 1;
          ans = (ans*ma1)*ma2;
          printf("%I64d\n",ans.m[1][1]);
    }
    return 0;
}