UVA 10655 Contemplation! Algebra(构造矩阵)

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本文详细解析了UVA10655代数问题，通过矩阵构造的方法解决给定a+b及ab值求特定表达式的值的问题。介绍了递推式和矩阵求解的具体步骤。

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UVA Problem 10655 Contemplation! Algebra

Accept: 0 Submit: 0
Time Limit: 3000 MS

Problem Description

Given the value of a+b and ab you will have to find the value of

Input

The input file contains several lines of inputs. Each line except the last line contains 3 non-negative integers p, q and n. Here p denotes the value of a+b and q denotes the value of ab. Input is terminated by a line containing only two zeroes. This line should not be processed. Each number in the input file fits in a signed 32-bit integer. There will be no such input so that you have to find the value of

Output

For each line of input except the last one produce one line of output. This line contains the value of . You can always assume that fits in a signed 64-bit integer.

Sample Input

Sample Output

   68
  
 91

Problem Idea

解题思路:

【题意】
给你a+b,ab,n的值,求

【类型】
矩阵构造

【分析】
首先,给你a+b,ab,可能第一想法是a,b为方程的两个根(韦达定理根深蒂固)

但是此题若是把a,b求解出来显然是不可取的

①方程不一定有实根

②就算是实根,精度不高

那么,换个思路,对进行化简,令f(n)=,则

就这样,递推式出来了,那么就可以转换成矩阵来求解

p=a+b;q=ab

特判n=0,n=1,n=2的情况,貌似n=2可以不用特判

n=0时,

n=1时,a+b=p

n=2时,(a+b)*(a+b)-2ab=p*p-2*q

另外,题目要求当且仅当输入的一行仅有0 0时,输入结束

故只判断p!=0&&q!=0是不对的,因为数据0 0 n(n≠0)是合法的

【时间复杂度&&优化】
O(logn)

题目链接→UVA 10655 Contemplation! Algebra

Source Code

/*Sherlock and Watson and Adler*/
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<complex>
#include<string>
#include<algorithm>
#include<iostream>
#define eps 1e-8
#define LL long long
#define bitnum(a) __builtin_popcount(a)
using namespace std;
const int N = 2;
const int M = 100005;
const int inf = 1000000007;
const int mod = 10;
typedef struct node
{
	long long a[N][N];
	void Init()
	{
		memset(a,0,sizeof(a));
		for(int i=0;i<N;i++)
			a[i][i]=1;
	}
}matrix;
matrix mul(matrix a,matrix b)//矩阵乘法
{
	matrix ans;
	for(int i=0;i<N;i++)
		for(int j=0;j<N;j++)
		{
			ans.a[i][j]=0;
			for(int k=0;k<N;k++)
				ans.a[i][j]+=a.a[i][k]*b.a[k][j];
		}
	return ans;
}
matrix add(matrix a,matrix b)//矩阵加法
{
	matrix ans;
	for(int i=0;i<N;i++)
		for(int j=0;j<N;j++)
			ans.a[i][j]=a.a[i][j]+b.a[i][j];
	return ans;
}
matrix pow(matrix a,int n)//求a^n
{
	matrix ans;
	ans.Init();
	while(n)
	{
		if(n%2)
			ans=mul(ans,a);
		n/=2;
		a=mul(a,a);
	}
	return ans;
}
int main()
{
    long long p,q,n;
    matrix ans,s;
    while(scanf("%lld%lld%lld",&p,&q,&n)==3)
    {
        if(n==0)
        {
            puts("2");
            continue;
        }
        if(n==1)
        {
            printf("%lld\n",p);
            continue;
        }
        if(n==2)
        {
            printf("%lld\n",p*p-2*q);
            continue;
        }
        s.a[0][0]=0;s.a[0][1]=1;
        s.a[1][0]=-q;s.a[1][1]=p;
        ans.a[0][0]=p;ans.a[0][1]=0;
        ans.a[1][0]=p*p-2*q;ans.a[1][1]=0;
        ans=mul(pow(s,n-1),ans);
        printf("%lld\n",ans.a[0][0]);
    }
    return 0;
}

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