HDU 1757-A Simple Math Problem(矩阵快速幂)

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A Simple Math Problem

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1757

Appoint description:

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 99991 1 1 1 1 1 1 1 1 120 5001 0 1 0 1 0 1 0 1 0

Sample Output

题意：就是两个式子
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); 求值。

思路：这算是初步学会如何构造矩阵了吧，其实也没什么难得，主要是想不好想，唉，还是自己太弱了。

x<10的时候好说，下面就说一下n>=10的情况。如下图

矩阵A 矩阵B

0 1 0 0 0 0 0 0 0 0 f0 f1

0 0 1 0 0 0 0 0 0 0 f1 f2

0 0 0 1 0 0 0 0 0 0 f2 f3

0 0 0 0 1 0 0 0 0 0 * f3 ----------> f4

0 0 0 0 0 1 0 0 0 0 f4 f5

0 0 0 0 0 0 1 0 0 0 f5 f6

0 0 0 0 0 0 0 1 0 0 f6 f7

0 0 0 0 0 0 0 0 1 0 f7 f8

0 0 0 0 0 0 0 0 0 1 f8 f9

a9 a8 a7 a6 a5 a4 a3 a2 a1 a0 f9 f10

我们看到规律了，每次要到下次个A*B，以此类推则由A*A*A.......A*B；

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>#include <set>#include <map>#include <queue>using namespace std;const int inf=0x3f3f3f3f;int mod;int a[20];struct node {    int mp[20][20];} init,res;struct node Mult(struct node x,struct node y){    struct node tmp;    int i,j,k;    for(i=0; i<10; i++)        for(j=0; j<10; j++) {            tmp.mp[i][j]=0;            for(k=0; k<10; 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<10; i++)        for(j=0; j<10; 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;};int main(){    int k,x,i,j;    while(~scanf("%d %d",&k,&mod)) {        if(k<10) {            printf("%d\n",k%mod);            continue ;        }        for(i=9; i>=0; i--)            a[i]=9-i;        for(i=0; i<10; i++) {            scanf("%d",&x);            init.mp[0][i]=x%mod;        }        for(i=1; i<10; i++) {            for(j=0; j<10; j++) {                if(i==j+1)                    init.mp[i][j]=1;                else                    init.mp[i][j]=0;            }        }        res=expo(init,k-9);        int ans=0;        for(j=0; j<10; j++) {            ans=(ans+res.mp[0][j]*a[j])%mod;        }        printf("%d\n",ans);    }    return 0;}

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