hdu1757 A Simple Math Problem(矩阵快速幂)

题目：

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3522    Accepted Submission(s): 2130

Problem Description

Lele now is thinking about a simple function f(x).

If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .

Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.

 

Input

The problem contains mutiple test cases.Please process to the end of file.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.

 

Output

For each case, output f(k) % m in one line.

 

Sample Input

  
  

   
   10 9999
1 1 1 1 1 1 1 1 1 1
20 500
1 0 1 0 1 0 1 0 1 0
  
  

 

Sample Output

  
  

   
   45
104
  
  

 

Author

linle

 

Source

2007省赛集训队练习赛（6）_linle专场

 

Recommend

lcy

题意：给一个函数的递推公式，求出它的第k项。

思路：矩阵快速幂。

代码：

#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cstdio>
#include <cmath>
#include<climits>
#include <algorithm>
#include <vector>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <fstream>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <list>
#include <stdexcept>
#include <functional>
#include <utility>
#include <ctime>
using namespace std;

#define PB push_back
#define MP make_pair

#define REP(i,x,n) for(int i=x;i<(n);++i)
#define FOR(i,l,h) for(int i=(l);i<=(h);++i)
#define FORD(i,h,l) for(int i=(h);i>=(l);--i)
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define RI(X) scanf("%d", &(X))
#define RII(X, Y) scanf("%d%d", &(X), &(Y))
#define RIII(X, Y, Z) scanf("%d%d%d", &(X), &(Y), &(Z))
#define DRI(X) int (X); scanf("%d", &X)
#define DRII(X, Y) int X, Y; scanf("%d%d", &X, &Y)
#define DRIII(X, Y, Z) int X, Y, Z; scanf("%d%d%d", &X, &Y, &Z)
#define OI(X) printf("%d",X);
#define RS(X) scanf("%s", (X))
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second
#define Swap(a, b) (a ^= b, b ^= a, a ^= b)
#define Dpoint  strcut node{int x,y}
#define cmpd int cmp(const int &a,const int &b){return a>b;}

 /*#ifdef HOME
    freopen("in.txt","r",stdin);
    #endif*/
const int MOD = 1e9+7;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<double> VD;
typedef long long LL;
typedef pair<int,int> PII;
//#define HOME

int Scan()
{
	int res = 0, ch, flag = 0;

	if((ch = getchar()) == '-')				//判断正负
		flag = 1;

	else if(ch >= '0' && ch <= '9')			//得到完整的数
		res = ch - '0';
	while((ch = getchar()) >= '0' && ch <= '9' )
		res = res * 10 + ch - '0';

	return flag ? -res : res;
}
/*----------------PLEASE-----DO-----NOT-----HACK-----ME--------------------*/

struct Matrix
{
    int m[15][15];
    void init()
    {
        MS0(m);
    }
};
Matrix mul(Matrix a,Matrix b,int M)
{
    Matrix c;
    c.init();
    for(int i=0;i<10;i++)
        for(int j=0;j<10;j++)
        for(int k=0;k<10;k++)
    {c.m[i][j]+=((long long )a.m[i][k]*b.m[k][j])%M;
    c.m[i][j]%=M;}
    return c;
}
Matrix mypow(Matrix a,int k,int M)
{
    Matrix ans;
    ans.init();
    for(int i=0;i<10;i++)
        ans.m[i][i]=1;
    Matrix temp=a;
    while(k)
    {
        if(k&1)
            ans=mul(ans,temp,M);
        k>>=1;
        temp=mul(temp,temp,M);

    }
    return ans;
}
int a[10];
int main()
{
int k,m;
while(RII(k,m)!=EOF)
{
    if(k<10)
    {
        printf("%d\n",k%m);
        continue;
    }
    for(int i=0;i<10;i++)
        RI(a[i]);
    Matrix A;
    A.init();
    for(int i=0;i<10;i++)
        A.m[0][i]=a[i];
    for(int i=1;i<10;i++)
        A.m[i][i-1]=1;
    Matrix res=mypow(A,k-9,m);
    int ans=0;
    for(int i=0;i<10;i++)
        ans=(ans+(long long)res.m[0][i]*(9-i))%m;
    printf("%d\n",ans);
}


        return 0;
}