A Simple Math Problem |
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
Total Submission(s): 155 Accepted Submission(s): 110 |
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:给你题目描述的函数,然后让你求f(k)%m的值 初步思路:简单的爆一下 #错误:超内存! #改进思路: 想一下这个式子的实质,不用递归写,用循环写,但是需要循环1e9次,肯定是不可行的,可以用矩阵相乘f(x)是从0,1,2...9乘上a0,a1,a2... .a9递推过来的这样就能得出来一个矩阵,线性代数还没学,矩阵真的有点麻烦,构造矩阵的时候老是想错了。 #再次错误:矩阵相乘虽然可以解决空间问题,但是没法解决时间问题;这里真的有点糊涂了,常数都有快速幂,矩阵的快速幂怎么就没有想起来呐! */ #include<bits/stdc++.h> #define ll long long using namespace std; ll m,k; struct Matrix{ ll m[12][12]; void init0(){//构造用来乘的矩阵(相当于“幺元”吧,随便想到了离散中的概念就这么叫吧) memset(m,0,sizeof m); for(int i=1;i<10;i++) m[i][i-1]=1; } void init1(){//构造最初的a0~a9 memset(m,0,sizeof m); for(int i=0;i<10;i++){ m[i][0]=9-i; } } }; Matrix Mul_Matrix(Matrix a,Matrix b){ Matrix c; c.init0(); for(int i=0;i<10;i++){ for(int j=0;j<10;j++){ c.m[i][j]=0; for(int k=0;k<10;k++){ c.m[i][j]+=(a.m[i][k]*b.m[k][j])%m; } c.m[i][j]%=m; } } return c; } /**************矩阵快速幂*********************/ Matrix Pow(Matrix a,Matrix b,int x){ while(x){ if(x&1){ b=Mul_Matrix(a,b); } a=Mul_Matrix(a,a); x>>=1; } return b; } /**************矩阵快速幂*********************/ Matrix a,b; int main(){ //freopen("in.txt","r",stdin); while(scanf("%lld%lld",&k,&m)!=EOF){ a.init0(); b.init1(); //cout<<k<<" "<<m<<endl; for(int i=0;i<10;i++) scanf("%lld",&a.m[0][i]); // for(int i=9;i<k;i++){ // b=Mul_Matrix(a,b); // for(int i=0;i<10;i++){ // for(int j=0;j<10;j++) // cout<<a.m[i][j]<<" "; // cout<<" "; // for(int j=0;j<10;j++) // cout<<b.m[i][j]<<" "; // cout<<endl; // } // } Matrix c=Pow(a,b,k-9);//利用矩阵快速幂 printf("%lld\n",c.m[0][0]); } return 0; }
A Simple Math Problem(矩阵快速幂)(寒假闭关第一题,有点曲折啊)
原文:http://www.cnblogs.com/wuwangchuxin0924/p/6308773.html