LOJ 1070 Algebraic Problem

题目链接：Light OJ 1070 Algebraic Problem

Problem Description

Given the value of a+b and ab you will have to find the value of an+bna^n+b^n. a and b not necessarily have to be real numbers.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains three non-negative integers, p, q and n. Here p denotes the value of a+b and q denotes the value of ab. 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 000^0.

Output

For each test case, print the case number and(an+bn) (a^n+b^n) modulo 2642^{64}.

Sample Input

2
10 16 2
7 12 3

Sample Output

Case 1: 68
Case 2: 91

题意：

给出p=a+b的值和q=ab的值，求(an+bn)(a^n+b^n)%2642^{64},(p,q,n都是32位有符号的数)

AC代码：

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
#include <iostream>
#include <locale>
#include <vector>
#include <string>
#include <iomanip>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <functional>using namespace std;int nnn=2;struct juzhen
{unsigned long long x[3][3];
}e,a,b,t;juzhen multi(juzhen A,juzhen B)
{juzhen c;memset(c.x,0,sizeof(c.x));for(int i=1; i<=nnn; i++)for(int j=1; j<=nnn; j++)for(int k=1; k<=nnn; k++)c.x[i][j]+=A.x[i][k]*B.x[k][j];return c;
}juzhen quick(unsigned long long nn)
{juzhen A=a,B=e;while(nn>0){if(nn&1)B=multi(B,A);nn>>=1;A=multi(A,A);}return B;
}int main()
{int T;scanf("%d",&T);for(int i=1; i<=T; i++){unsigned long long p,q,n;scanf("%llu%llu%llu",&p,&q,&n);memset(e.x,0,sizeof(e.x));memset(a.x,0,sizeof(a.x));memset(b.x,0,sizeof(b.x));for(int j=1;j<=nnn;j++)e.x[j][j]=1;a.x[1][1] = p;a.x[1][2] = -q;a.x[2][1] = 1;b.x[1][1] = p;b.x[2][1] = 2;if(!n){printf("Case %d: 2\n",i);continue;}t=quick(n-1);t=multi(t,b);printf("Case %d: %llu\n",i,t.x[1][1]);}return 0;
}

LOJ 1070 Algebraic Problem相关推荐

LightOJ 1070 Algebraic Problem （推导+矩阵快速幂）
题目链接:LightOJ 1070 Algebraic Problem 题意:已知a+b和ab的值求a^n+b^n.结果模2^64. 思路: 1.找递推式得到递推式之后就是矩阵快速幂了注意:模2^ ...
[LightOJ1070]Algebraic Problem
题目:Algebraic Problem 链接:https://vjudge.net/problem/LightOJ-1070 分析: 1)$ a^n+b^n = ( a^{n-1}+b^{n-1} ...
LIGHT OJ1070 Algebraic Problem
题目: Given the value of a+b and ab you will have to find the value of a^n+b^n. a and b not necessaril ...
Loj 6036 「雅礼集训 2017 Day4」编码 - 2-sat
题目传送门唯一的传送门题目大意给定$n$个串,每个串只包含 '0','1','?' ,其中 '?' 至多在每个串中出现1次,它可以被替换为 '0' 或 '1' .问是否可能任意两个不同的串不满足 ...
当代数学大师迈克尔阿蒂亚证明黎曼猜想
去9月24日,89岁高龄的当代数学大师迈克尔阿蒂亚先生亲自前往海德堡高级数学论坛公开证明黎曼猜想,此举吸引了全球数学界的目光. 阿蒂亚指出:黎曼猜想的核心问题是素数分布定理,两者关系密切.国人对此事件 ...
list下界_下界理论
list下界下界理论 (Lower Bound Theory) Lower bound (L(n)) is a property of the specific problem i.e. the s ...
LOJ 2721 「NOI2018」屠龙勇士——扩展中国剩余定理
题目:https://loj.ac/problem/2721 1.注意别一输入 p[ i ] 就 a[ i ] %= p[ i ] ,因为在 multiset 里找的时候还需要真实值. 2.注意用 m ...
Loj #6287 诗歌
link: https://loj.ac/problem/6287 一开始差点写FFT了23333,并且FFT还能算这样的三元组的数量而且还不用要求这是一个排列.... 但这太大材小用了(而且很可能被 ...
Loj #2036. 「SHOI2015」自动刷题机
link : https://loj.ac/problem/2036 这个显然具有单调性,N小的话更容易A题,不仅因为A一次题减少的代码,并且A题的下限也低. 所以直接上二分就行了,注意上限一定不要设 ...

LOJ 1070 Algebraic Problem

LOJ 1070 Algebraic Problem

题目链接：Light OJ 1070 Algebraic Problem

Problem Description

Input

Output

Sample Input

Sample Output

题意：

AC代码：

LOJ 1070 Algebraic Problem相关推荐

最新文章

热门文章