[HDU4990] 等比公式+取模运算公式 或矩阵求递推法~

Reading comprehension

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 384    Accepted Submission(s): 185

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

 

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打个表，递推公式 a[n]=a[n-1]+2*a[n-2]+1

建立矩阵 [a[n-1],a[n-2],1]*[1,1,0]

                                              [2,0,0]

                                              [1,0,1]

【a[2],a[1],1】* 后面矩阵的 n-2次幂 n=1,2的时候特判就行

注意爆int

#define DeBUG#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <string>#include <set>#include <sstream>#include <map>#include <list>#include <bitset>using namespace std ;#define zero {0}#define INF 0x3f3f3f3f#define EPS 1e-6#define TRUE true#define FALSE falsetypedef long long LL;const double PI = acos(-1.0);//#pragma comment(linker, "/STACK:102400000,102400000")inline int sgn(double x){    return fabs(x) < EPS ? 0 : (x < 0 ? -1 : 1);}#define N 100005const int MAXN = 5;long long mod;struct Matrix{    long long mat[MAXN][MAXN];    void Zero()    {        memset(mat, 0, sizeof(mat));    }    void Unit()    {        memset(mat, 0, sizeof(mat));        for (int i = 0; i < MAXN; i++)            mat[i][i] = 1;    }    void output()    {        for (int i = 0; i < MAXN; i++)        {            for (int j = 0; j < MAXN; j++)            {                printf("%d ", mat[i][j]);            }            printf("\n");        }    }};Matrix operator*(Matrix &a, Matrix &b){    Matrix tmp;    tmp.Zero();    for (int k = 0; k < MAXN; k++)    {        for (int i = 0; i < MAXN; i++)        {            if (!a.mat[i][k])                continue;            for (int j = 0; j < MAXN; j++)            {                tmp.mat[i][j] += a.mat[i][k] * b.mat[k][j] % mod;                if ( tmp.mat[i][j] >= mod)                    tmp.mat[i][j] -= mod;            }        }    }    return tmp;}Matrix operator ^(Matrix a, int k){    Matrix tmp;    tmp.Unit();    if(k<=1)        return a;    for (; k; k >>= 1)    {        if (k & 1)            tmp = tmp * a;        a = a * a;    }    return tmp;}int main(){#ifdef DeBUGs    freopen("C:\\Users\\Sky\\Desktop\\1.in", "r", stdin);#endif    Matrix mt;    long long n,m;    while (scanf("%I64d%I64d", &n, &m) + 1)    {        mod=m;        if(n==1)        {            printf("%I64d\n", 1%m);            continue;        }        else if(n==2)        {            printf("%I64d\n", 2%m);            continue;        }        mt.Zero();        mt.mat[0][0] = 1;        mt.mat[0][1] = 1;        mt.mat[1][0] = 2;        mt.mat[2][0] = 1;        mt.mat[2][2] = 1;        mt = mt ^ (n-2);        Matrix mt2;        mt2.Zero();        mt2.mat[0][0] = 2;        mt2.mat[0][1] = 1;        mt2.mat[0][2] = 1;        Matrix ans = mt2 * mt;        printf("%I64d\n", ans.mat[0][0]);    }    return 0;}

推公式，公式看程序就懂了，两个等比数列其实 关键求(4^dm -1)/3 %mod

利用公式a/b%m = a%(b*m)/b注意这时候再取模是%（b*m）快速幂的时候得变了，因为这个逗B了。。。

#define DeBUG#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <vector>#include <stack>#include <queue>#include <string>#include <set>#include <sstream>#include <map>#include <list>#include <bitset>using namespace std ;#define zero {0}#define INF 0x3f3f3f3f#define EPS 1e-6#define TRUE true#define FALSE falsetypedef long long LL;const double PI = acos(-1.0);//#pragma comment(linker, "/STACK:102400000,102400000")inline int sgn(double x){    return fabs(x) < EPS ? 0 : (x < 0 ? -1 : 1);}#define N 100005long long MOD;LL pow2(LL a, LL b, LL mod){    LL r = 1, base = a;    while (b != 0)    {        if (b % 2)            r = r * base % mod;        base = base * base % mod;        b /= 2;    }    return r;}int main(){#ifdef DeBUGs    freopen("C:\\Users\\Sky\\Desktop\\1.in", "r", stdin);#endif    long long n, m;    while (scanf("%I64d%I64d", &n, &m) + 1)    {        MOD = m;        int dm;        if (n & 1)            dm = (n - 1) / 2 + 1;        else            dm = (n - 2) / 2 + 1;        long long ans = (pow2(4, dm, (3 * MOD)) - 1) % (3 * MOD) / 3;        if (n % 2 == 0)        {            ans = (ans * 2) % MOD;        }        printf("%I64d\n", ans);    }    return 0;}