ICPC2017网络赛（沈阳）1005&HDU6198 number （矩阵+快速幂）

最新推荐文章于 2019-03-18 20:48:45 发布

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最新推荐文章于 2019-03-18 20:48:45 发布

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分类专栏： ACM：数学+概率论+数论+组合数学文章标签：矩阵递推

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number number number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 26 Accepted Submission(s): 13

Problem Description

We define a sequence

F :

⋅

F0=0,F1=1 ;

⋅

Fn=Fn−1+Fn−2 (n≥2) .

Give you an integer

k , if a positive number

n can be expressed by

n=Fa1+Fa2+...+Fak where

0≤a1≤a2≤⋯≤ak , this positive number is

mjf−good . Otherwise, this positive number is

mjf−bad .
Now, give you an integer

k , you task is to find the minimal positive

mjf−bad number.
The answer may be too large. Please print the answer modulo 998244353.

Input

There are about 500 test cases, end up with EOF.
Each test case includes an integer

k which is described above. (

1≤k≤109 )

Output

For each case, output the minimal

mjf−bad number mod 998244353.

Sample Input

Sample Output

Source

2017 ACM/ICPC Asia Regional Shenyang Online

【题意】：

给出一个数k，问用k个斐波那契数相加，得不到的数最小是几。

【解析】：

暴力写了下前几项，发现有规律。f[n] = f[n-1]*3 - f[n-2] + 1

然后用矩阵快速幂可以直接得出答案。

【代码】

#include <stdio.h>
#include <stdlib.h>  
#include <string.h>  
#include <iostream>  
#include <algorithm> 
#define mset(a,i) memset(a,i,sizeof(a))
using namespace std;
typedef long long ll;
const int mod=998244353;
ll qpow(ll n,ll m){n%=mod;ll ans=1;while(m){if(m%2) 
		ans=(ans*n)%mod;m/=2;n=(n*n)%mod;}return ans;}
struct zhen{  
    int r,c;//行列始于1  
    ll a[16][16];  
    zhen(int R=0,int C=0){  
        memset(a,0,sizeof(a));  
        r=R;c=C;  
    }  
};   
zhen operator*(zhen A,zhen B)//矩阵A*B  
{  
    zhen C(A.r,B.c);  
    for(int i=1;i<=A.r;i++)  
        for(int j=1;j<=B.c;j++)  
            for(int k=1;k<=A.c;k++){  
                C.a[i][j]+=((A.a[i][k]+mod)*(mod+B.a[k][j]))%mod;
                C.a[i][j]%=mod;
            }  
    return C;  
}  
zhen qpow(zhen A,ll m)//方阵A的m次幂  
{  
    zhen ans(A.r,A.c);  
    for(int i=1;i<=A.r;i++) ans.a[i][i]=1;//单位矩阵  
    while(m)  
    {  
        if(m%2)ans=ans*A;  
        A=A*A;  
        m/=2;  
    }  
    return ans;  
}
int main()  
{  
    ll n;  
    while(cin>>n)
	{
	    if(n==1){  
	        puts("4");continue; 
	    }  
	    if(n==2){
	    	puts("12");continue;
		}
	    zhen A(3,3);//构造矩阵A
	    A.a[1][1]=3;  
	    A.a[1][3]=A.a[2][1]=A.a[3][3]=1;
		A.a[1][2]=-1;
	    //矩阵A构造完毕  
	    //所以 Xn+1 = A^(n-2) * X3;
	    zhen X3(A.r,1),X;
	    X3.a[1][1]=12;
	    X3.a[2][1]=4;
	    X3.a[3][1]=1;
	    X=qpow(A,n-2)*X3;
	    ll ans=X.a[1][1];
	    cout<<ans%mod<<endl;  
	}
}

前几项打表的代码：

#include <stdio.h>
#include <string.h>  
using namespace std;
typedef long long ll;
int dp[200][2000];
int main()
{
	ll c[1100]={0,1,1};
	for(int i=2;i<1020;i++)
		c[i]=(c[i-1]+c[i-2]);
	memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    for (int i = 0; i <= 40; i++)     //枚举总类
    {
        for (int num = 1; num <= 40; num++)    //枚举个数
        {
            for (int j = c[i]; j <= 1000; j++)      //枚举容量
            {

                dp[num][j] += dp[num - 1][j - c[i]];
            }
        }
    }
	for(int i=0;i<=40;i++)
		for(int j=1;j<=1000;j++)
			if(dp[i][j]==0)
			{
				printf("%d\n",j);break;
			}
	return 0;
}