HDU 4990 Reading comprehension（递推+快速幂 或 矩阵快速幂）

Reading comprehension

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 930    Accepted Submission(s): 362

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

 

Source

BestCoder Round #8

 

题目大意：如题中程序所示，给出n和m，快速求出ans的值。

解题思路：

1、递推+快速幂

当i是偶数时，a[i]=2*a[i-1]（i>=2）；

当i是奇数时，a[i]=2*a[i-1]+1（i>=3，a[1]=1）。

对于偶数项，可以得到递推公式为：a[i]=2*a[i-1]=2*（2*a[i-2]+1）=4*a[i-2]+2（i为偶数，i>=4，a[2]=2）

如果令a[0]=0；则a[i]=4*a[i-2]+2（i为偶数，i>=2，a[0]=0）则

a[2]=4*a[0]+2；

a[4]=4*a[2]+2=4^2*a[0]+8+2；

a[6]=4*a[4]+2=4^2*a[2]+8+2=4^3*a[0]+32+8+2；

a[8]=4*a[6]+2=4^2*a[4]+8+2=4^3*a[2]+32+8+2=4^4*a[0]+128+32+8+2；

……………………………………………………………………………………；

a[n]=4^(n/2)*a[0]+2+8+……+2*4^(n/2-1)=0+2+8+……+2*4^(n/2-1)=2*(1-4^(n/2))/(1-4)=(2^(n+1)-2)/3（n>=2n为偶数）；

对于奇数项，利用递推公式a[i]=2*a[i-1]+1（i>=3，a[1]=1），因为i为奇数时，i-1一定是偶数。

a[n]=2*a[n-1]+1=2*(2^n-2)/3+1=(2^(n+1)-4)/3+1=(2^(n+1)-1)/3（n>=1，n为奇数）；

对于(a / b) mod c有(a / b) mod c = a mod (b*c) / b成立。

证明：设x ≡ a/b (mod c)，根据同余式的性质两边同时乘以b，得bx ≡ a (mod b*c)，

根据同余式的性质得a ≡ bx (mod b*c)，所以x=a mod (b*c) / b。

综上所述：

当n为偶数时，a[n] = (2^(n+1) - 2)/3；当n为奇数时，a[n] = (2^(n+1) - 1)/3

快速幂求解即可。

代码如下：

#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <string>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <numeric>#include <iomanip>#include <bitset>#include <sstream>#include <fstream>#include <limits.h>#define debug "output for debug\n"#define pi (acos(-1.0))#define eps (1e-6)#define inf (1<<28)#define sqr(x) (x) * (x)#define mod 1000000007using namespace std;typedef long long ll;typedef unsigned long long ULL;ll POW(ll a,ll b,ll m){    ll ret=1;    while(b)    {        if(b&1)            ret=ret*a%m;        a=a*a%m;        b>>=1;    }    return ret;}int main(){    ll n,m,ans;    while(scanf("%I64d%I64d",&n,&m)!=EOF)    {        if(n&1)        {            ans=(POW(2,n+1,3*m)-1)/3;            printf("%I64d\n",ans);        }        else        {            ans=(POW(2,n+1,3*m)-2)/3;            printf("%I64d\n",ans);        }    }    return 0;}

2、矩阵快速幂

当i是偶数时，a[i]=2*a[i-1]（i>=2）；

当i是奇数时，a[i]=2*a[i-1]+1（i>=3，a[1]=1）。

进一步可以得到：当i是偶数时，a[i]=2*a[i-1]=4*a[i-2]+2，如果令a[0]=0，则

当i是奇数时，先计算i-1（因为i-1是偶数），再利用a[i]=2*a[i-1]+1。

代码如下：

#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <string>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <numeric>#include <iomanip>#include <bitset>#include <sstream>#include <fstream>#include <limits.h>#define debug "output for debug\n"#define pi (acos(-1.0))#define eps (1e-6)#define inf (1<<28)#define sqr(x) (x) * (x)using namespace std;typedef long long ll;typedef unsigned long long ULL;#define MAX 4ll n,m,mod;struct Matr{    ll w[MAX][MAX];    Matr()    {        memset(w,0,sizeof(w));    }};Matr operator *(Matr a,Matr b){    Matr c;    ll i,j,k;    for(k=1;k<=2;k++)    {        for(i=1;i<=2;i++)        {            if(a.w[i][k]==0)                continue;            for(j=1;j<=2;j++)            {                if(b.w[k][j]==0)                    continue;                c.w[i][j]=(c.w[i][j]+a.w[i][k]*b.w[k][j])%mod;            }        }    }    return c;}Matr operator ^(Matr a,ll k){    Matr c;    ll i,j;    for(i=1;i<=2;i++)        c.w[i][i]=1;    while(k)    {        if(k&1)            c=c*a;        a=a*a;        k>>=1;    }    return c;}int main(){    ll i,j,k;    Matr a,b,c;    while(~scanf("%I64d%I64d",&n,&m))    {        a.w[1][1]=4;a.w[1][2]=1;        a.w[2][1]=0;a.w[2][2]=1;        b.w[1][1]=0;        b.w[2][1]=2;        mod=m;        if(n&1)        {            c=a^((n-1)/2);            printf("%I64d\n",(c.w[1][2]*2*2+1)%mod);        }        else        {            c=a^(n/2);            printf("%I64d\n",c.w[1][2]*2%mod);        }    }    return 0;}