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HDU 1588-Gauss Fibonacci(矩陣快速冪+二分求矩陣和)

編輯：C++入門知識

HDU 1588-Gauss Fibonacci(矩陣快速冪+二分求矩陣和)

Gauss Fibonacci Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1588 Appoint description:

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 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.

Output

For each line input, out the value described above.

Sample Input

 2 1 4 100
2 0 4 100

Sample Output

 21
12

這道題和poj3233的解題思路差不多，不過這道題需要自己推出規律來，還是太弱了，我怎麼也認為輸出的和應該是左上角的值，結果怎麼也不對，最後去看了題解，看到輸出的是右上角，然後看了人家寫的題解思路，瞬間懂了。sadddddd

點擊打開鏈接(附上巨巨博客)。用於構造斐波那契的矩陣為1 1

1 0

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int inf=0x3f3f3f3f;
long long mod;
struct node
{
    long long mp[3][3];
}init,res1,res2,res;
struct node Mult(struct node x,struct node y)
{
    int i,j,k;
    struct node tmp;
    for(i=0;i<2;i++)
    for(j=0;j<2;j++){
        tmp.mp[i][j]=0;
        for(k=0;k<2;k++){
            tmp.mp[i][j]=(tmp.mp[i][j]+x.mp[i][k]*y.mp[k][j])%mod;
        }
    }
    return tmp;
};
struct node expo(struct node x,int k)
{
    struct node tmp;
    int i,j;
    for(i=0;i<2;i++)
    for(j=0;j<2;j++){
        if(i==j)
            tmp.mp[i][j]=1;
        else
            tmp.mp[i][j]=0;
    }
    while(k){
        if(k&1) tmp=Mult(tmp,x);
        x=Mult(x,x);
        k>>=1;
    }
    return tmp;
};
struct node add(struct node x,struct node y)
{
    struct node tmp;
    int i,j;
    for(i=0;i<2;i++)
        for(j=0;j<2;j++)
        tmp.mp[i][j]=(x.mp[i][j]+y.mp[i][j])%mod;
    return tmp;
};
struct node sum(struct node x,int k)
{
    struct node tmp;
    if(k==1) return x;
    if(k&1) return add(sum(x,k-1),expo(x,k));
    tmp=sum(x,k/2);
    return add(tmp,Mult(tmp,expo(x,k/2)));
};
int main()
{
    int k,b,n;
    while(~scanf("%d %d %d %lld",&k,&b,&n,&mod)){
        init.mp[0][0]=1;
        init.mp[0][1]=1;
        init.mp[1][0]=1;
        init.mp[1][1]=0;
        res1=expo(init,b);
        res2=expo(init,k);
        res=add(res1,Mult(res1,sum(res2,n-1)));
        printf("%lld\n",res.mp[0][1]);
    }
    return 0;
}