hdu4546 So Easy! 矩阵乘法

So Easy!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 899 Accepted Submission(s): 244

Problem Description

　　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!

Input

　　There are several test cases, each test case in one line contains four positive integers: a, b, n, m. Where 0< a, m < 2¹⁵, (a-1)²< b < a², 0 < b, n < 2³¹.The input will finish with the end of file.

Output

　　For each the case, output an integer S_n.

Sample Input

2 3 1 20132 3 2 20132 2 1 2013

Sample Output

4144

Source

2013 ACM-ICPC长沙赛区全国邀请赛——题目重现

Recommend

zhoujiaqi2010

这可一点都不so easy啊，注意取模的时候可能为负!

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;__int64 mod;struct matrix {    __int64 a[2][2];    matrix operator * (matrix num)const    {        int i,j,k;        matrix temp;        for(i=0;i<2;i++)        {            for(j=0;j<2;j++)            {                temp.a[i][j]=0;                for(k=0;k<2;k++)                {                    temp.a[i][j]+=a[i][k]*num.a[k][j]%mod;                    temp.a[i][j]%=mod;                }            }        }        return temp;    }};int main(){   __int64 a,b,n,m;   matrix start,original,temp;   while(scanf("%I64d%I64d%I64d%I64d",&a,&b,&n,&m)!=EOF)   {       mod=m;      if(n==1)      {          printf("%I64d\n",2*(a%m)%m);          continue;      }      else if(n==2)      {          printf("%I64d\n",(2*(a%m)*(a%m)+2*(b%m)%m));          continue;      }      n=n-2;      a=a%m,b=b%m;      start.a[0][0]=2*a%m;      start.a[0][1]=1;      start.a[1][0]=(b-a*a%m+m)%m;      start.a[1][1]=0;      original.a[0][0]=1,original.a[1][1]=1;      original.a[1][0]=0,original.a[0][1]=0;      temp.a[0][0]=(2*a*a%m+2*b%m)%m;temp.a[0][1]=2*a%m;      temp.a[1][0]=0,temp.a[1][1]=0;      while(n)      {          if(n&1)            original=original*start;          n=n>>1;          start=start*start;      }      temp=temp*original;      printf("%I64d\n",temp.a[0][0]%m);   }    return 0;}