HDU1588解题报告

Gauss Fibonacci

                                        Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                   Total Submission(s): 2539    Accepted Submission(s): 1054

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.

 

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 1002 0 4 100

 

Sample Output

2112

 

Author

DYGG

 

Source

HDU “Valentines Day” Open Programming Contest 2007-02-14

 

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             参考代码：

#include<cstdio>#include<iostream>#include<cmath>#include<cstring>#include<algorithm>#include<string>#include<vector>#include<map>#include<set>#include<stack>#include<queue>#include<ctime>#include<cstdlib>#include<iomanip>#include<utility>#define pb push_back#define mp make_pair#define CLR(x) memset(x,0,sizeof(x))#define _CLR(x) memset(x,-1,sizeof(x))#define REP(i,n) for(int i=0;i<n;i++)#define Debug(x) cout<<#x<<"="<<x<<" "<<endl#define REP(i,l,r) for(int i=l;i<=r;i++)#define rep(i,l,r) for(int i=l;i<r;i++)#define RREP(i,l,r) for(int i=l;i>=r;i--)#define rrep(i,l,r) for(int i=1;i>r;i--)#define read(x) scanf("%d",&x)#define put(x) printf("%d\n",x)#define ll long long#define lson l,m,rt<<1#define rson m+1,r,rt<<11using namespace std;int k,b,n,M;struct mat{    ll d[3][3];  //必须用long long，否则会爆int} A,B,C,E;mat multi(mat a,mat b){    mat ans;    rep(i,0,2)    {        rep(j,0,2)        {            ans.d[i][j]=0;            rep(k,0,2)                ans.d[i][j]+=a.d[i][k]*b.d[k][j];            ans.d[i][j]%=M;        }    }    return ans;}mat add(mat a,mat b){    mat ans;    rep(i,0,2)        rep(j,0,2)           ans.d[i][j]=(a.d[i][j]+b.d[i][j])%M;    return ans;}mat quickmulti(mat A,int n){    if(n==0) return E;    if(n==1) return A;    mat ans=E;    while(n)    {        if(n&1)        {            n--;            ans=multi(ans,A);        }        else        {            n>>=1;            A=multi(A,A);        }    }    return ans;}mat binarymulti(mat A,int n)   //二分矩阵等比之和{    if(n==1) return A;    mat tmp=binarymulti(A,n/2),tmp1=quickmulti(A,n/2);    mat ans=add(tmp,multi(tmp,tmp1));    if(n&1) ans=add(ans,multi(multi(tmp1,tmp1),A));  //multi(multi(tmp1,tmp1),A)即A^n    return ans;}int main(){    E.d[0][0]=E.d[1][1]=A.d[0][0]=A.d[0][1]=A.d[1][0]=B.d[0][0]=1;    E.d[0][1]=E.d[1][0]=A.d[1][1]=B.d[0][1]=B.d[1][0]=B.d[1][1]=0;    C.d[0][0]=C.d[0][1]=C.d[1][1]=0,C.d[1][0]=1;    //E表示单位矩阵，A表示斐波那契数列的矩阵，B矩阵是包含F(1)和F(0)的列向量，C矩阵相当于是包含F(0)和F(-1)的矩阵，C矩阵主要针对b=0的情况。                  while(~scanf("%d%d%d%d",&k,&b,&n,&M))    {        mat mt1=B,mt2=quickmulti(A,k); //mt2表示A^k        if(b==0) mt1=C;        else mt1=multi(quickmulti(A,b-1),mt1); //计算出包含F(b)和F(b-1)的列向量        mt2=binarymulti(mt2,n-1);  //计算A+A^2+……+A^(n-1)的和矩阵        mt2=add(mt2,E);  //加上A的0次幂，即单位矩阵        mt1=multi(mt2,mt1); //乘以包含F(b)和F(b-1)的列向量        printf("%I64d\n",mt1.d[0][0]%M);    }}