HDU 5451 Best Solver

最新推荐文章于 2021-04-13 22:00:04 发布

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最新推荐文章于 2021-04-13 22:00:04 发布

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分类专栏： HDU ----矩阵乘法文章标签： HDU

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----矩阵乘法

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本文探讨了一种用于解决特定数学表达式问题的高效算法，该算法利用循环节判断和矩阵乘法来简化计算过程。通过实例演示，展示了如何在给定整数x和质数M的情况下，计算表达式(y=(5+26√)^1+2x)的整数部分除以M后的结果。此方法特别适用于需要大量计算和优化场景。

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Problem Description

The so-called best problem solver can easily solve this problem, with his/her childhood sweetheart.

It is known that

y=(5+26√)1+2x.
For a given integer

x (0≤x<232) and a given prime number

M (M≤46337), print

[y]%M. (

[y] means the integer part of

Input

An integer

T (1<T≤1000), indicating there are

T test cases.
Following are

T lines, each containing two integers

x and

M, as introduced above.

Output

The output contains exactly

T lines.
Each line contains an integer representing

[y]%M.

Sample Input

Sample Output

Case #1: 97
Case #2: 969
Case #3: 16537
Case #4: 969
Case #5: 40453
Case #6: 10211
Case #7: 17947

矩阵乘法+循环节判断
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
#include<iostream>
using namespace std;
typedef long long LL;
int T,t=0,m,f[50000];
unsigned int n;

struct martix
{
    unsigned int a[2][2];
    martix(){memset(a,0,sizeof(a));}
    martix operator*(const martix &b)
    {
        martix c;
        for (int i=0;i<2;i++)
            for (int j=0;j<2;j++)
                for (int k=0;k<2;k++)
                    (c.a[i][k]+=a[i][j]*b.a[j][k])%=m;
        return c;
    }
};

bool operator!=(const martix &a,const martix &b)
{
    if (a.a[0][0]!=b.a[0][0]) return true;
    
    if (a.a[0][1]!=b.a[0][1]) return true;
    
    if (a.a[1][0]!=b.a[1][0]) return true;
    return false;
}

int get(unsigned int n,int m)
{
    int ans=1;
    for (int i=2;n;n>>=1)
    {
        if (n&1) (ans*=i)%=m;
        i=(i*i)%m;
    }
    return (ans+1)%m;
}

martix mget(martix a,int x)
{
    martix c=a;
    if (x<=0) return c;
    for (x-=1;x;x>>=1)
    {
        if (x&1) c=c*a;
        a=a*a;
    }
    return c;
}

int query()
{
    if (f[m]) return f[m];
        martix b,c;
        int loop=1;
        b.a[0][0]=10%m;
        b.a[0][1]=1%m;
        b.a[1][0]=(m-1)%m;
        c=b*b;
        while (b!=c) {c=c*b; loop++; }
        f[m]=loop;
        return loop;
}

int main()
{
    memset(f,0,sizeof(f));
    scanf("%d",&T);
    while (T--)
    {
        scanf("%u%d",&n,&m);
        martix a,b,c;
        int loop=1;
        a.a[0][0]=98%m;
        a.a[0][1]=10%m;
        b.a[0][0]=10%m;
        b.a[0][1]=1%m;
        b.a[1][0]=(m-1)%m;
        n=get(n,query());
        c=mget(b,n-1);
        a=a*c; 
        printf("Case #%d: %d\n",++t,(a.a[0][1]+m-1)%m);
    }
    return 0;
}