LightOJ 1070 Algebraic Problem

Given the value of a+b and ab you will have to find the value of aⁿ+bⁿ. 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 qdenotes 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 0⁰.

Output

For each test case, print the case number and (aⁿ+bⁿ) modulo 2⁶⁴.

Sample Input

2

10 16 2

7 12 3

Sample Output

Case 1: 68

Case 2: 91

嗯，一开始说对2^64取模，一脸懵逼，心想这个怎么做啊。然后用的是unsigned long long  超过64会自动取模的。。。。emmmmmm那你如果是个负数也取模成正数么。

估计应该是的，负数应该变成正的，实际上也不能算错吧。

这个咋一看会想到(a+b)^n然后，展开发现仍旧不能做，实际上我们换个想法，能不能找到递推式呢？不难发现a^(n+1) + b^(n+1) = p(a^n+b^n) - q(a^(n-1) + b^(n-1)。

也就是要求的F(n) = p*F(n-1) - q*F(n-2)。

然后就可以矩阵快速幂加速了。

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;typedef unsigned long long LL;const LL mod = 7;const int Size = 2;long long n,p,q;struct matrix{    LL a[4][4]= {{0}};    matrix operator *(const matrix b)const    {        matrix ans;        for(int i = 0; i < Size; ++i)            for(int k = 0; k < Size; ++k)                for(int j = 0; j < Size; ++j)                {                    ans.a[i][j] = (ans.a[i][j] + a[i][k]*b.a[k][j]);                }        return ans;    }};LL get_ans(){    if(n == 0)return 2;    matrix ans,base;    ans.a[0][0] = p;    ans.a[0][1] = 2;    base.a[0][0] = p;    base.a[0][1] = 1;    base.a[1][0] = -q;    n--;    while(n)    {        if(n & 1)ans = ans*base;        base = base*base;        n >>= 1;    }    return ans.a[0][0];}int main(){    //cout <<INT_MAX<<endl;    int t;    int ca = 0;    scanf("%d",&t);    while(t--)    {        scanf("%lld%lld%lld",&p,&q,&n);        printf("Case %d: %llu\n",++ca,get_ans());    }    return 0;}