HDU 2256 Problem of Precision (矩阵快速幂+ 化简 +向下取整)

Problem of Precision

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1565    Accepted Submission(s): 951

Problem Description

 

Input

The first line of input gives the number of cases, T. T test cases follow, each on a separate line. Each test case contains one positive integer n. (1 <= n <= 10^9)

 

Output

For each input case, you should output the answer in one line.

 

Sample Input

3125

 

Sample Output

997841

 

代码如下:

#include <iostream>#include <cstring>#include <queue>#include <algorithm>#include <stdio.h>#include <cmath>#include <string>#include <stack>typedef long long ll;const int MOD=1024;const int N =20;const int MAXN=2;using namespace std;struct Matrix{ll arr[N][N];void init(){memset(arr,0,sizeof(arr));for(int i=0;i<MAXN;i++)arr[i][i]=1;//初始化}void iinit(){    memset(arr,0,sizeof(arr));    arr[0][0]=arr[1][1]=5;    arr[0][1]=12;    arr[1][0]=2;}}A;Matrix mul(Matrix X,Matrix Y)// 矩阵乘法{Matrix ans;for(int i=0;i<MAXN;i++)for(int j=0;j<MAXN;j++){ans.arr[i][j]=0;for(int k=0;k<MAXN;k++){ans.arr[i][j]+=X.arr[i][k]*Y.arr[k][j];ans.arr[i][j]%=MOD;}}return ans;}Matrix Q_pow(Matrix B,int n)// 矩阵快速幂{Matrix ans;ans.init();while(n){if(n&1)ans=mul(ans,B);n>>=1;B=mul(B,B);}return ans;}Matrix Add(Matrix a,Matrix b)  //(a+b)%mod 矩阵加法{    int i,j,k;    Matrix ans;    for(i=0;i<MAXN;i++)        for(j=0;j<MAXN;j++)        {            ans.arr[i][j]=a.arr[i][j]+b.arr[i][j];            ans.arr[i][j]%=MOD;        }    return ans;}Matrix Sum(Matrix a,int n)// 矩阵和{int m;Matrix ans,pre;if(n==1) return ans;m=n/2;pre=Sum(a,m);ans=Add(pre,mul(pre,Q_pow(a,m)));if(n&1)ans=Add(ans,Q_pow(a,n));return ans;}int main(){    int T;    cin>>T;    int n;    while(T--)    {        scanf("%d",&n);        Matrix ans;        ans.iinit();        if(n==1)        {            printf("9\n");            continue;        }        ans=Q_pow(ans,n-1);        ll res=0;        //cout<<ans.arr[0][0]<<" "<<ans.arr[0][1]<<" "<<endl;        res=(ans.arr[0][0]*5+ans.arr[0][1]*2)%MOD;        printf("%lld\n",(2*res-1)%MOD);    }    return 0;}