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首先我们看g=k*i+b; 是一个等差数列
如果能推出f(g)这个函数也是一个等差或者等比数列，就可以得出一个公式
f(k*i+b) 0<=i<n
建立一个二分矩阵A=[1 1,1 0]
f(b)=A^b
则:
f(b)=A^b
f(k+b)=A^k+b
f(2*k+b)=A^2*k+b.
.
.
f((n-1)*k+b)=A^(n-1)*k+b
我们就可以得出一个等比数列:
首项：A^b
公比：A^k
项数：n
(res是单位矩阵)
运用等比数列求和公式得出：sum=A^b*(res+A^k+(A^k)^2+(A^k)^3+...+(A^k)^(n-1))
需要注意的一点是：当b=0的情况
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例如:A+A^2+A^3+A^4+A^5+A^6=(A+A^2+A^3)+A^3*(A+A^2+A^3) 用递归写
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Gauss Fibonacci
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1657 Accepted Submission(s): 722
Problem Description
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.
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.
Input
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.
Each of them will not exceed 1,000,000,000.
Output
For each line input, out the value described above.
Sample Input
2 1 4 100
2 0 4 100
2 0 4 100
Sample Output
21
12
12
Author
DYGG
Source
Recommend
linle
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
long long int k,b,n,MOD;
struct Matrix
{
long long m[3][3];
};
Matrix res;
Matrix mut(Matrix a,Matrix b)
{
Matrix c;
memset(c.m,0,sizeof(c.m));
for(int i=1;i<=2;i++)
{
for(int j=1;j<=2;j++)
{
for(int k=1;k<=2;k++)
{
c.m[j]=((a.m[k]*b.m[k][j])%MOD+c.m[j])%MOD;
}
}
}
return c;
}
Matrix quickpow(Matrix a,int k)
{
Matrix res;
memset(res.m,0,sizeof(res.m));
res.m[1][1]=res.m[2][2]=1;
while(k>1)
{
if(k&1)
{
k--;
res=mut(res,a);
}
else
{
k/=2;
a=mut(a,a);
}
}
return mut(a,res);
}
Matrix add(Matrix a,Matrix b)
{
Matrix c;
for(int i=1;i<=2;i++)
{
for(int j=1;j<=2;j++)
c.m[j]=(a.m[j]+b.m[j])%MOD;
}
return c;
}
Matrix sum(Matrix p,int k)
{
if(k==1) return p;
else if(k&1) return (add(quickpow(p,k),sum(p,k-1)));
else return (mut(sum(p,k/2),add(res,quickpow(p,k/2))));
}
int main()
{
Matrix a,ak,ab,ans;
a.m[1][1]=a.m[1][2]=a.m[2][1]=1;
a.m[2][2]=0;
res.m[1][1]=res.m[2][2]=1;
res.m[1][2]=res.m[2][1]=0;
while(cin>>k>>b>>n>>MOD)
{
ak=quickpow(a,k);
if(b!=0)
ab=quickpow(a,b);
else
ab=res;//0次幂当然是单位阵了
ans=add(res,sum(ak,n-1));
ans=mut(ab,ans);
printf("%I64d\n",ans.m[2][1]);
}
return 0;
}