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

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

1 103 100

 

Sample Output

15

 

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<<"---"<<endltypedef 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;}