矩阵两道题目codeforces 450B 和hdu 5015

最新推荐文章于 2020-02-13 21:30:31 发布

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本文探讨了233矩阵的构造与求解方法，通过递归与矩阵运算解决复杂序列计算问题，同时分享了在算法实现过程中的细节处理与优化技巧。

233 matrix

In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it would be 233, 2333, 23333... (it means a 0,1 = 233,a 0,2 = 2333,a 0,3 = 23333...) Besides, in 233 matrix, we got a i,j = a i-1,j +a i,j-1( i,j ≠ 0). Now you have known a 1,0,a 2,0,...,a n,0, could you tell me a n,m in the 233 matrix?

Input

There are multiple test cases. Please process till EOF.

For each case, the first line contains two postive integers n,m(n ≤ 10,m ≤ 10 9). The second line contains n integers, a 1,0,a 2,0,...,a n,0(0 ≤ a i,0 < 2 31).

Output

For each case, output a n,m mod 10000007.

Sample Input

Sample Output

234
2799
72937

Hint

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int modn=1e7+7;
const int maxn=15;
#define ll long long
ll n,m;
struct node{
   ll s[maxn][maxn];
   node(){
       memset(s,0,sizeof(s));
   }
};
node multi(node a,node b){
   node t;
   for(int i=0;i<n;i++){
       for(int j=0;j<n;j++){
           for(int k=0;k<n;k++){
               t.s[i][j]=(t.s[i][j]+a.s[i][k]*b.s[k][j]%modn+modn)%modn;
           }
       }
   }
   return t;
}
node quick(node a,ll nn){
   node res;
   for(int i=0;i<n;i++){
       res.s[i][i]=1;
   }
   while(nn){
       if(nn&1) res=multi(res,a);
       a=multi(a,a);
       nn>>=1;
   }
   return res;

}
int main(){
   while(scanf("%lld%lld",&n,&m)!=EOF){
       ll a[15]; node b;
       for(int i=1;i<=n;i++){
       scanf("%lld",&a[i]);
       }
       a[0]=23;
       a[n+1]=3;
       n++;
       n++;
       for(int i=0;i<n;i++){
           if(i!=n-1) b.s[i][0]=10;
           for(int j=1;j<=i;j++){
               if(i!=n-1)
               b.s[i][j]=1;
           }
           b.s[i][n-1]=1;
       }
/*       for(int i=0;i<n;i++){
           for(int j=0;j<n;j++){
               cout<<b.s[i][j]<<" ";
           }
           cout<<endl;
       }*/
       node res=quick(b,m);
       //b[n-1][n-1]=1;
       ll ans=0;
   /*       for(int i=0;i<n;i++){
           for(int j=0;j<n;j++){
               cout<<res.s[i][j]<<" h";
           }
           cout<<endl;
       }*/
       for(int i=0;i<n;i++){
           ans=(a[i]*(res.s[n-2][i])%modn+ans)%modn;
       //   cout<<a[i]<<" "<<res.s[n-2][i]<<"wer"<<endl;
       }
       cout<<ans<<endl;
   }
   return 0;
}

这道题目，怎么说呢，递归的话，很有可能用矩阵，我直接找规律，又忘记了复杂度过不了了

还有细节问题，快速幂，稍微推一下就行了

jzzhu and sequences

Jzzhu has invented a kind of sequences, they meet the following property:

You are given x and y, please calculate fn modulo 1000000007 (109 + 7).

Input

The first line contains two integers x and y (|x|, |y| ≤ 109). The second line contains a single integer n (1 ≤ n ≤ 2·109).

Output

Output a single integer representing fn modulo 1000000007 (109 + 7).

Examples

Input

2 3
3

Output

Input

0 -1
2

Output

1000000006

Note

In the first sample, f2 = f1 + f3, 3 = 2 + f3, f3 = 1.

In the second sample, f2 = - 1; - 1 modulo (109 + 7) equals (109 + 6).

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
const int modn=1e9+7;
struct node{
	ll m[2][2];
	node(){
		memset(m,0,sizeof(m));
	}
};
node multi(node a,node b){
	node res;
	for(int i=0;i<2;i++){
		for(int j=0;j<2;j++) 
		res.m[i][j]=0;
	}
	for(int i=0;i<2;i++){
		for(int j=0;j<2;j++){
			for(int k=0;k<2;k++){
				res.m[i][j]=(ll)((ll)(a.m[i][k]*b.m[k][j]+modn)+res.m[i][j])%modn;
			//	cout<<a.m[i][k]<<" "<<b.m[k][j]<<" "<<res.m[i][j]<<" "<<i<<" "<<j<<"end"<<endl;
			}
		}
	}
//	cout<<res.m[0][0]<<" "<<res.m[0][1]<<" "<<res.m[1][0]<<"f "<<res.m[1][1]<<endl;
	return res;
}
//node res;
node  qui(node r,int n){
	node res;
	for(int i=0;i<2;i++){
		for(int j=0;j<2;j++) 
		res.m[i][j]=0;
	}
	for(int i=0;i<2;i++){
		res.m[i][i]=1;
	}
	res.m[0][1]=0;
	res.m[1][0]=0;
	while(n){
	//	cout<<n<<"res"<<endl;
		if(n&1)
		res=multi(res,r); 
	//	cout<<res.m[0][1]<<"ans"<<endl;
		r=multi(r,r);
		n>>=1;
	}
	
	return res;
}
//ll a[2][2];
int main(){
	ll x,y,n;
while(	scanf("%lld%lld%lld",&x,&y,&n)!=EOF)
{

	node a;
	node b;
	b.m[0][0]=x;
	b.m[0][1]=y;
	a.m[0][1]=-1;
	a.m[1][0]=1;
	a.m[1][1]=1; 
	a.m[0][0]=0;	
	if(n==2){
		cout<<(y%modn+modn)%modn<<endl;
		return 0;
	}
	if(n==1){
		cout<<(x%modn+modn)%modn<<endl;
		return 0;
	}
	
	node res=qui(a,n-2);
	//res=multi(b,res);
	ll ans=(ll)((ll)(res.m[0][1]*x)%modn+(ll)(res.m[1][1]*y)%modn)%modn;
	
	if(ans<0) ans+=modn; 
	if(ans<0) ans+=modn;
	cout<<ans<<endl;
//	cout<<(res.m[0][1]+modn)%modn<<endl;
}
	return 0;
}

这道题目忘了用longlong了一直wa, 差点把我劝退了。

总之，细节问题，很扎心