【CUGBACM15级BC第8场 B】hdu 4990 Reading comprehension

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本文通过分析一道程序阅读理解题目，给出了求解该问题的两种方法，并详细解释了每种方法的具体实现步骤及核心思想。

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Reading comprehension

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2072 Accepted Submission(s): 827

Problem Description

Read the program below carefully then answer the question.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include<iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include<vector>

const int MAX=100000*2;
const int INF=1e9;

int main()
{
  int n,m,ans,i;
  while(scanf("%d%d",&n,&m)!=EOF)
  {
    ans=0;
    for(i=1;i<=n;i++)
    {
      if(i&1)ans=(ans*2+1)%m;
      else ans=ans*2%m;
    }
    printf("%d\n",ans);
  }
  return 0;
}

Input

Multi test cases，each line will contain two integers n and m. Process to end of file.
[Technical Specification]
1<=n, m <= 1000000000

Output

For each case，output an integer，represents the output of above program.

Sample Input

1 10
3 100

Sample Output

1
5

思路：

先用源程序跑几个数出来！1， 2， 5， 10， 21， 42 好了，再用这几个数字找他们的通项式

#include <iostream>
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <bitset>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <algorithm>
#include <functional>
#define PI acos(-1)
#define eps 1e-8
#define inf 0x3f3f3f3f
#define debug(x) cout<<"---"<<x<<"---"<<endl
typedef long long ll;


ll mod;
struct matrix
{
    ll mat[5][5];
    matrix()
    {
        memset(mat, 0, sizeof(mat));
    } int R, C;
    matrix operator*(matrix other);
};

matrix matrix::operator*(matrix other)
{
    matrix c;
    for (int i = 1; i <= R; i++) for (int j = 1; j <= C; j++) for (int k = 1; k <= C; k++)
            {
                c.mat[i][j] = (c.mat[i][j] + mat[i][k] * other.mat[k][j] % mod) % mod;
            }
    c.R = R;
    c.C = other.C;
    return c;
}
matrix fast_mod(matrix base, int k)
{
    matrix tmp;
    tmp.R = tmp.C = 3;
    for (int i = 1; i <= 3; i++)
    {
        tmp.mat[i][i] = 1;
    }
    while (k)
    {
        if (k & 1)
        {
            tmp = tmp * base;
        }
        base = base * base;
        k >>= 1;
    }
    return tmp;
}

int main()
{
    int n;
    while (~scanf("%d%I64d", &n, &mod))
    {
        matrix ans, base;
        ans.R = 1;
        ans.C = 1;
        base.R = base.C = 3;
        if (n == 1)
        {
            printf("%d\n", 1 % mod);
        }
        else if (n == 2)
        {
            printf("%d\n", 2 % mod);
        }
        else
        {
            ans.R = 1;
            ans.C = 3;
            ans.mat[1][1] = 1;
            ans.mat[1][2] = 2;
            ans.mat[1][3] = 1;
            base.mat[2][3] = base.mat[1][1] = base.mat[1][2] = base.mat[1][3] = 0;
            base.mat[2][1] = 2;
            base.mat[2][2] = 4;
            base.mat[3][1] = base.mat[3][3] = 1;
            base.mat[3][2] = 2;
            int k = (n - 2) / 2;
            if (n & 1)
            {
                base = fast_mod(base, k + 1);
                ans = ans * base;
                printf("%I64d\n", ans.mat[1][1]);
            }
            else
            {
                base = fast_mod(base, k);
                ans = ans * base;
                printf("%I64d\n", ans.mat[1][2]);
            }
        }
    }
    return 0;
}