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; }