HDU 4686 Arc of Dream（构造矩阵快速幂）

最新推荐文章于 2021-08-04 15:15:41 发布

Mr_Treeeee

最新推荐文章于 2021-08-04 15:15:41 发布

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分类专栏： HDU 矩阵乘法

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本文介绍了一种通过构造矩阵的方法来高效计算弧梦曲线在特定点的取值，并提供了一个完整的C++代码实现。该方法适用于处理大规模数据集，尤其当输入参数N非常大时，能够快速准确地计算出弧梦曲线的值。

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Arc of Dream

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 5059 Accepted Submission(s): 1584

Problem Description

An Arc of Dream is a curve defined by following function:

where
a ₀ = A0
a _i = a _i-1*AX+AY
b ₀ = B0
b _i = b _i-1*BX+BY
What is the value of AoD(N) modulo 1,000,000,007?

Input

There are multiple test cases. Process to the End of File.
Each test case contains 7 nonnegative integers as follows:
N
A0 AX AY
B0 BX BY
N is no more than 10 ¹⁸, and all the other integers are no more than 2×10 ⁹.

Output

For each test case, output AoD(N) modulo 1,000,000,007.

Sample Input

Sample Output

Author

Zejun Wu (watashi)

Source

2013 Multi-University Training Contest 9

点击打开链接

这题就是构造矩阵，链接里的和我的没差，就是顺序不太一样。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define  LL long long
const LL N = 5;
const LL p = 1e9+7;
struct mx
{
    LL a[N+1][N+1];
};
mx cheng(mx a,mx b)
{
    mx ans;
    for(int i=1;i<=N;i++)
    {
        for(int j=1;j<=N;j++)
        {
            ans.a[i][j]=0;
            for(int k=1;k<=N;k++)
            {
                (ans.a[i][j]+=a.a[i][k]*b.a[k][j])%=p;
            }
        }
    }
    return ans;
}
mx qkm(mx base,LL mi)
{
    mx ans;
    for(LL i=1;i<=N;i++)
    {
        for(LL j=1;j<=N;j++)
        {
            ans.a[i][j]=i==j;
        }
    }
    while(mi)
    {
        if(mi&1) ans=cheng(ans,base);
        base=cheng(base,base);
        mi>>=1;
    }
    return ans;
}
int main()
{
    LL n;
    while(~scanf("%lld",&n))
    {
        LL a0,ax,ay;
        LL b0,bx,by;
        scanf("%lld %lld %lld %lld %lld %lld",&a0,&ax,&ay,&b0,&bx,&by);
        LL x1=ax;
        LL x2=bx;
        LL y1=ay;
        LL y2=by;
        if(n==0) printf("0\n");
        else if(n==1) printf("%lld\n",a0*b0%p);
        else
        {
            mx base;
            base.a[1][1]=1;base.a[1][2]=0;      base.a[1][3]=0; base.a[1][4]=0; base.a[1][5]=0;
            base.a[2][1]=1;base.a[2][2]=x1*x2%p;base.a[2][3]=0; base.a[2][4]=0; base.a[2][5]=0;
            base.a[3][1]=0;base.a[3][2]=y1*x2%p;base.a[3][3]=x2;base.a[3][4]=0; base.a[3][5]=0;
            base.a[4][1]=0;base.a[4][2]=y2*x1%p;base.a[4][3]=0; base.a[4][4]=x1;base.a[4][5]=0;
            base.a[5][1]=0;base.a[5][2]=y1*y2%p;base.a[5][3]=y2;base.a[5][4]=y1;base.a[5][5]=1;
            base=qkm(base,n-1);
            LL a1=(x1*a0%p+y1)%p;
            LL b1=(x2*b0%p+y2)%p;
            LL b2=(x2*b1%p+y2)%p;
            LL a2=(x1*a1%p+y1)%p;
          //  printf("%lld %lld\n",a2,b2);
            LL z[6];
            z[1]=a0*b0%p;
            z[2]=a1*b1%p;
            z[3]=b1;
            z[4]=a1;
            z[5]=1;
            LL ans=0;
            for(int k=1;k<=5;k++)
            {
                ans+=z[k]*base.a[k][1];
            }
            printf("%lld\n",ans%p);
        }
    }
}