uva 10655 - Contemplation! Algebra (构造矩阵,快速幂)

uva10655题目链接

题意:

给出a+b和ab的值,问a^n+b^n

思路:

构造矩阵,手写一下很显然...

转移矩阵M=[0 , 1]

[-q,p ]

初始矩阵M1=[p           ]

[p^2-2*q]

快速幂即可。

有个坑点在于..读入的结束是p=0&q=0,并且只有这两个输入。

如果用p=0&&q=0作为终止条件,那么就会将三个输入,但p==0&&q==0的情况错误得终止...

正确的做法是 while (~scanf("%lld%lld%lld",&p,&q,&n)==3)

/* ***********************************************
Author :111qqz
Created Time :2017年10月01日 星期日 03时01分38秒
File Name :10655.cpp
************************************************ */
 1#include <cstdio>
 2#include <cstring>
 3#include <iostream>
 4#include <algorithm>
 5#include <vector>
 6#include <queue>
 7#include <set>
 8#include <map>
 9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define PB push_back
14#define fst first
15#define sec second
16#define lson l,m,rt<<1
17#define rson m+1,r,rt<<1|1
18#define ms(a,x) memset(a,x,sizeof(a))
19typedef long long LL;
20#define pi pair < int ,int >
21#define MP make_pair
 1using namespace std;
 2const double eps = 1E-8;
 3const int dx4[4]={1,0,0,-1};
 4const int dy4[4]={0,-1,1,0};
 5const int inf = 0x3f3f3f3f;
 6LL p,q,n;
 7struct Mat
 8{
 9    LL mat[105][105];
10    void clear()
11    {
12    ms(mat,0);
13    }
14}M,M1;
15Mat operator * (Mat a,Mat b)
16{
17    Mat c;
18    c.clear();
19    for ( int i = 0 ; i < 2 ; i++)
20    for ( int j = 0 ; j < 2 ; j++)
21        for ( int k = 0 ; k < 2 ; k++)
22        c.mat[i][j] += a.mat[i][k] * b.mat[k][j];
23    return c;
24}
25Mat operator ^ (Mat a,LL b)
26{
27    Mat ret;
28    ret.clear();
29    for ( int i = 0 ; i < 2 ; i++) ret.mat[i][i] = 1LL;
 1    while (b>0)
 2    {
 3    if (b&1) ret = ret * a;
 4    a = a * a;
 5    b = b >> 1LL;
 6    }
 7    return ret;
 8}
 9LL solve()
10{
11    if (n==0) return 2LL;
12    if (n==1) return p;
13    if (n==2) return p*p-2*q;
14    M1.clear();
15    M1.mat[0][0] = p;
16    M1.mat[1][0] = p*p-2*q;
17    M.clear();
18    M.mat[0][1] = 1;
19    M.mat[1][0] = -q;
20    M.mat[1][1] = p;
21    Mat ans;
22    ans.clear();
23    ans = (M^(n-2))*M1;
24    return ans.mat[1][0];
25}
26int main()
27{
28    #ifndef  ONLINE_JUDGE 
29//  freopen("./in.txt","r",stdin);
30  #endif
31    while (scanf("%lld%lld%lld",&p,&q,&n)==3)
32    {
33        //if (p==0&&q==0) break;
34        //scanf("%lld",&n);
35        LL ans = solve();
36        printf("%lld\n",ans);
37    }
1  #ifndef ONLINE_JUDGE  
2  fclose(stdin);
3  #endif
4    return 0;
5}