东大OJ-1040-Count-快速幂方法求解斐波那契-

Many ACM team name may be very funny,such as "Complier_Error","VVVVV".Oh,wait for a minute here.Is it "W W"+"V",or "W"+"V V V",or something others we can treat as?There are several ways we can treat this name "VVVVV" (5 'V's),asV V can be treat as a W.For 5 'V's,our have 8 ways.They are:V V V V VV W WW W VV W V VW V WW V V VV V W VV V V WThe problem here is that for n 'V's,how many ways do we have to treat it?Becausethe answer may be too large, you should output the answer module by p.(Ifn is 0,then we have just one way.)

输入

There are multiple test cases. The first line of the input contains an integerM, meaning the number of the test cases.For each test cases, there aretwo integers nand pin a single line.You can assume that0<=n<=2100000000,0<p<=2009.

输出

For each test case, output the answer with case number in a single line.

样例输入

2
5 5
4 7

样例输出

3

#include<iostream>using namespace std;struct m{int a[2][2]; };m  mul(m a, m b,int p){m c;int i, j,k;for (i = 0; i < 2;i++)for (j = 0; j < 2; j++){c.a[i][j] = 0;for (k = 0; k < 2; k++)c.a[i][j] +=( (a.a[i][k] %p)* (b.a[j][k]%p))%p;c.a[i][j] %= p;}return c;}m go(m a, int n,int p){if (n == 1)return a;m b = go(a, n / 2, p);m c = mul(b, b, p);if (n % 2 == 1)c = mul(c, a, p);return c;}int main(){//freopen("in.txt", "r", stdin);int t;cin >> t;m a = { 0, 1, 1, 1 };while (t-- > 0){int n, p;cin >> n >> p;if (n == 0){ cout << 1 << endl; continue; }cout << go(a, n, p).a[1][1]<<endl;}return 0;}

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