HDU：Gauss Fibonacci（矩阵快速幂+二分）

http://acm.hdu.edu.cn/showproblem.php?pid=1588

[align=left]Problem Description[/align]
Without expecting, Angel replied quickly.She says: "I'v heard that you'r a very clever boy. So if you wanna me be your GF, you should solve the problem called GF~. " How good an opportunity that Gardon can not give up! The "Problem GF" told by Angel is actually "Gauss Fibonacci". As we know ,Gauss is the famous mathematician who worked out the sum from 1 to 100 very quickly, and Fibonacci is the crazy man who invented some numbers.
Arithmetic progression: g(i)=k*i+b; We assume k and b are both non-nagetive integers.
Fibonacci Numbers: f(0)=0 f(1)=1 f(n)=f(n-1)+f(n-2) (n>=2)
The Gauss Fibonacci problem is described as follows: Given k,b,n ,calculate the sum of every f(g(i)) for 0<=i<n The answer may be very large, so you should divide this answer by M and just output the remainder instead.

[align=left]Input[/align]
The input contains serveral lines. For each line there are four non-nagetive integers: k,b,n,M Each of them will not exceed 1,000,000,000.

[align=left]Output[/align]
For each line input, out the value described above.

[align=left]Sample Input[/align]

2 1 4 100

2 0 4 100

[align=left]Sample Output[/align]

21
12

题目解析：

用于构造斐波那契的矩阵为

0,1

1,1

设这个矩阵为A。

sum=f(b)+f(k+b)+f(2*k+b)+f(3*k+b)+........+f((n-1)*k+b)

<=>sum=A^b+A^(k+b)+A^(2*k+b)+A^(3*k+b)+........+A^((n-1)*k+b)

<=>sum=A^b+A^b*(A^k+A^2*k+A^3*k+.......+A^((n-1)*k))（1）

设矩阵B为A^k;

那么（1）式为

sum=A^b+A^b*(B+B^2+B^3+......+B^(n-1));

显然，这时候就可以用二分矩阵做了，括号内的就跟POJ 3233的形式一样了。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#define inf 0x3f3f3f3f
#define LL __int64//int就WA了
using namespace std;
struct ma
{
LL a[2][2];
} init,res,B,C;
int mod,k,b,n,K;
void Init()
{
init.a[0][0]=0;
init.a[0][1]=1;
init.a[1][0]=1;
init.a[1][1]=1;
}
ma Mult(ma x,ma y)
{
ma tmp;
for(int i=0; i<2; i++)
{
for(int j=0; j<2; j++)
{
tmp.a[i][j]=0;
for(int z=0; z<2; z++)
{
tmp.a[i][j]=(tmp.a[i][j]+x.a[i][z]*y.a[z][j])%mod;
}
}
}
return tmp;
}
ma Pow(ma x,int K)
{
ma tmp;
for(int i=0; i<2; i++)
{
for(int j=0; j<2; j++)
tmp.a[i][j]=(i==j);
}
while(K!=0)
{
if(K&1)
tmp=Mult(tmp,x);
K>>=1;
x=Mult(x,x);
}
return tmp;
}
ma Add(ma x,ma y)
{
ma tmp;
for(int i=0; i<2; i++)
{
for(int j=0; j<2; j++)
{
tmp.a[i][j]=(x.a[i][j]+y.a[i][j])%mod;
}
}
return tmp;
}
ma Sum(ma x,int K)
{
ma tmp,y;
if(K==1)
return x;
tmp=Sum(x,K/2);
if(K&1)
{
y=Pow(x,K/2+1);
tmp=Add(Mult(y,tmp),tmp);
tmp=Add(tmp,y);
}
else
{
y=Pow(x,K/2);
tmp=Add(Mult(y,tmp),tmp);
}
return tmp;
}
/*另外一种写法
matrix Sum(matrix x, int k)
{
if(k==1) return x;
if(k&1)
return Add(Sum(x,k-1),Pow(x,k));  //如果k是奇数，求x^k+sum(x,k-1)
matrix tmp;
tmp=Sum(x,k>>1);
return Add(tmp,Mult(tmp,Pow(x,k>>1)));
}
*/
int main()
{
while(scanf("%d%d%d%d",&k,&b,&n,&mod)!=EOF)
{
Init();
B=Pow(init,k);
C=Pow(init,b);
res=Sum(B,n-1);
res=Mult(C,res);
res=Add(C,res);
printf("%I64d\n",res.a[1][0]);
}
return 0;
}