Contemplation! Algebra
Given the value of a+b and ab you will have to find the value of a
n + b
n
Input
The input file contains several lines of inputs. Each line except the last line contains 3 non-negative
integers p, q and n. Here p denotes the value of a+b and q denotes the value of ab. Input is terminated
by a line containing only two zeroes. This line should not be processed. 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 line of input except the last one produce one line of output. This line contains the value of
a
n + b
n. You can always assume that a
n + b
n fits in a signed 64-bit integer.
Sample Input
10 16 2
7 12 3
0 0
Sample Output
68
91
附：debug：
input：
0 0 0
0 0 1
0 0 200000000
1 3 0
999999 999999 0
1 0 200000000
2 0 61
2 1 0
2 1 100
2 1 200000000
0 0 0
0 1 1
0 1 2
0 1 3
0 1 199999999
0 1 200000000
0 1 200000001
0 1 200000002
5 6 7
10 20 10
90 87 3
999 231 2
0 0
output：
2 0 0 2 2 1 2305843009213693952 2 2 2 2 0 -2 0 0 2 0 -2 2315 393600000 705510 997539 题意：输入p,q,n;p=a+b，q=a*b，输出a^n+b^nTIP:令f（n）=a^n+b^n;则f(n）*(a+b)=a^(n+1)+b^(n+1)+a*b^n+b*a^n =f(n+1)+a*b*f(n-1) 即：f（n+1)=-(a+b)*f(n)+a*b*f(n-1)好了，矩阵上场，对于形如f（n+1)=p*f(n)+q*f(n-1)的；都可以转换为（p,q） (1,0)…………%…………这是个2*2的矩阵
然后{f（n）,f(n-1)}……（此为1*2的矩阵）=(那个矩阵）^(n-2)*｛f(2),f(1)｝;
注意：如果输入0,0,2；你的程序运行不？
scanf（……………………==3），我的如果不写的话就超时。
代码:
1 ///a^(n+1)+b^(n+1)=(a^n+b^n)p−(a^(n−1)+b^(n−1))
2 #include<string.h>
3 #include<stdio.h>
4
5 long long p,q,n;
6
7 struct mat
8 {
9 long long v[2][2];
10 mat(){memset(v,0,sizeof(v));}
11 mat operator * (mat c)
12 {
13 mat ans;
14 for(int i=0;i<2;i++)///矩阵乘法再加上原有的原数
15 {
16 for(int j=0;j<2;j++)
17 {
18 for(int k=0;k<2;k++)
19 {
20 ans.v[i][j]=ans.v[i][j]+v[i][k]*c.v[k][j];
21 }
22 }
23 }
24 return ans;
25 }
26 };
27
28 mat pow_mod(mat x,long long k)
29 {
30 mat ans;
31 ans.v[0][0]=ans.v[1][1]=1;
32
33 while(k)///二分求定以后的*
34 {
35 if(k&1)
36 ans=ans*x;
37 x=x*x;
38 k>>=1;
39 }
40
41 return ans;
42 }
43
44 long long solve()
45 {
46 if(n==0)
47 return 2;
48 if(n==1)
49 return p;
50 if(n==2)
51 return p*p-2*q;
52
53 mat a;
54 a.v[0][0]=p;
55 a.v[0][1]=-q;
56 a.v[1][0]=1;
57 a=pow_mod(a,n-2);
58
59 return a.v[0][1]*p+a.v[0][0]*(p*p-2*q);
60 }
61
62 int main()
63 {
64 while(scanf("%lld%lld%lld",&p,&q,&n)==3)
65 {
66 printf("%lld\n",solve());
67 }
68 return 0;
69 }