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Yet another Number Sequence
Let’s define another number sequence, given by the following function:
f(0) = a
f(1) = b
f(n) = f(n − 1) + f(n − 2), n > 1
When a = 0 and b = 1, this sequence gives the Fibonacci Sequence. Changing the values of a and
b, you can get many different sequences. Given the values of a, b, you have to find the last m digits of
f(n).
Input
The first line gives the number of test cases, which is less than 10001. Each test case consists of a
single line containing the integers a b n m. The values of a and b range in [0, 100], value of n ranges in
[0, 1000000000] and value of m ranges in [1, 4].
Output
For each test case, print the last m digits of f(n). However, you should NOT print any leading zero.
Sample Input
4
0 1 11 3
0 1 42 4
0 1 22 4
0 1 21 4
Sample Output
89
4296
7711
946
题意：给出a,b,n,m;a为第一个数，b为第二个数，然后以a，b为前两项a[n+1]=a[n]+a[n-1];构造数列
m为去几位数，即%10的几次方;n为此数列的第几个数
代码：
1 #include<iostream> 2 #include<cstdio> 3 4 using namespace std; 5 6 const int mod[5]={0,10,100,1000,10000}; 7 8 #define ll long long 9 10 ll a,b,m,n; 11 12 struct matrix 13 { 14 int map[2][2]; 15 }ans,base; 16 17 matrix unit={1,0,0,1}; 18 19 matrix multi(matrix a,matrix b) 20 { 21 matrix c; 22 for(int i=0;i<2;i++) 23 { 24 for(int j=0;j<2;j++) 25 { 26 c.map[i][j]=0; 27 for(int k=0l;k<2;k++) 28 { 29 c.map[i][j]+=(a.map[i][k]*b.map[k][j])%mod[m]; 30 } 31 c.map[i][j]%=mod[m]; 32 } 33 } 34 return c; 35 } 36 37 void pow(int n) 38 { 39 base.map[0][0]=base.map[0][1]=base.map[1][0]=1; 40 base.map[1][1]=0; 41 ans.map[0][0]=ans.map[1][1]=1; 42 ans.map[0][1]=ans.map[1][0]=0; 43 while(n) 44 { 45 if(n&1) 46 ans=multi(ans,base); 47 base=multi(base,base); 48 n>>=1; 49 } 50 } 51 52 int main() 53 { 54 int t; 55 scanf("%d",&t); 56 while(t--) 57 { 58 scanf("%lld%lld%lld%lld",&a,&b,&n,&m); 59 if(n==0) 60 return a; 61 else if(n==1) 62 return b; 63 else 64 { 65 pow(n-1); 66 ll result=(ans.map[0][0]*b+ans.map[0][1]*a)%mod[m]; 67 printf("%lld\n",result); 68 } 69 } 70 return 0; 71 }
//很有代表性
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原文地址：http://www.cnblogs.com/moqitianliang/p/4742334.html
