HDU - 1757 A Simple Math Problem （矩阵快速幂）

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.
OutputFor 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

思路：矩阵太大，用5*5的代为表达吧~

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define debug(a,i) cout<<#a<<"["<<i<<"] = "<<a[i]<<endl;
#define ls (t<<1)
#define rs ((t<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100086;
const int maxm = 100086;
const int inf = 2.1e9;
const ll Inf = 999999999999999999;
int mod;
const double eps = 1e-6;
const double pi = acos(-1);struct Matrix{int mp[15][15];
};
Matrix mul(Matrix a,Matrix b,int n){Matrix ans;for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){ans.mp[i][j]=0;for(int k=1;k<=n;k++){ans.mp[i][j]+=a.mp[i][k]*b.mp[k][j];}ans.mp[i][j]%=mod;}}return ans;
}Matrix q_pow(Matrix a,int b,int n){Matrix ans;memset(ans.mp,0,sizeof(ans.mp));for(int i=1;i<=n;i++){ans.mp[i][i]=1;}while (b){if(b&1){ans=mul(ans,a,n);}b>>=1;a=mul(a,a,n);}return ans;
}int main()
{
//    ios::sync_with_stdio(false);
//    freopen("in.txt","r",stdin);
Matrix exa;int n;while(scanf("%d%d",&n,&mod)!=EOF){memset(exa.mp,0,sizeof(exa.mp));for(int i=1;i<=10;i++){scanf("%d",&exa.mp[1][i]);}if(n<10){printf("%d\n",n);continue;}for(int i=2;i<=10;i++){exa.mp[i][i-1]=1;}exa = q_pow(exa,n-9,10);ll ans=0;for(int i=1;i<=10;i++){ans+=(10-i)*exa.mp[1][i];//注意这里是10-ians%=mod;}printf("%lld\n",ans);}return 0;
}

View Code

转载于:https://www.cnblogs.com/ZGQblogs/p/10880578.html