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