注意题目的一个关键条件(a-1)2< b < a2 ， 于是可以知道 0 < a-√b < 1 ,所以 (a-√b)^n < 1 . 然后 (a+ √b)^n+(a-√b)^n 的值为整数且正好是 (a+√b)^n的向上取整.
然后就可以得到递推式 f[n+1]=2*a*f[n]-(a*a-b)*f[n-1] .
然后构造矩阵。 幂乘即可.
So Easy!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1364 Accepted Submission(s): 424
Problem Description
A sequence Sn is defined as:
Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate Sn.
You, a top coder, say: So easy!
Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate Sn.
You, a top coder, say: So easy!
Input
There are several test cases, each test case in one line contains four positive integers: a, b, n, m. Where 0< a, m < 215, (a-1)2< b < a2, 0 < b, n < 231.The input will finish with the end of file.
Output
For each the case, output an integer Sn.
Sample Input
2 3 1 2013
2 3 2 2013
2 2 1 2013
Sample Output
4
14
4
Source
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zhoujiaqi2010
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <string.h> 4 #include <algorithm> 5 #include <math.h> 6 #include <map> 7 #include <queue> 8 #include <sstream> 9 #include <iostream> 10 using namespace std; 11 #define INF 0x3fffffff 12 13 typedef __int64 LL; 14 15 LL a,b,n,m; 16 LL g[2][2]; 17 LL MOD; 18 19 void cal(LL s[2][2],LL t[2][2]) 20 { 21 LL tmp[2][2]; 22 memset(tmp,0,sizeof(tmp)); 23 for(int k=0;k<2;k++) 24 for(int i=0;i<2;i++) 25 for(int j=0;j<2;j++) 26 { 27 tmp[i][j]=(tmp[i][j]+s[i][k]*t[k][j])%MOD; 28 } 29 for(int i=0;i<2;i++) 30 for(int j=0;j<2;j++) 31 s[i][j]=tmp[i][j]; 32 } 33 34 35 int main() 36 { 37 //freopen("C:\Users\Administrator\Desktop\in.txt","r",stdin); 38 //freopen("C:\Users\Administrator\Desktop\in.txt","w",stdout); 39 while(scanf("%d%d%d%d",&a,&b,&n,&m)!=EOF) 40 { 41 MOD=m; 42 g[0][0]=2*a; g[0][1]=1; 43 g[1][0]=b-a*a; g[1][1]=0; 44 45 n--; 46 LL sum[2][2]; 47 for(int i=0;i<2;i++) 48 for(int j=0;j<2;j++) 49 if(i==j) sum[i][j]=1; 50 else sum[i][j]=0; 51 while(n) 52 { 53 if( (n&1)!=0) 54 cal(sum,g); 55 cal(g,g); 56 n>>=1; 57 } 58 LL ans=((2*a*sum[0][0]+2*sum[1][0])%MOD+MOD)%MOD; 59 cout<<ans<<endl; 60 } 61 return 0; 62 }