lightoj 1006 矩阵快速幂

素颜马尾好姑娘i 2022-07-28 01:09 100阅读 0赞

1006 - Hex-a-bonacci

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   <td>Time Limit: <span style="color:#B45F04">0.5 second(s)</span></td> 
   <td>Memory Limit: <span style="color:#B45F04">32 MB</span></td> 
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Given a code (not optimized), and necessary inputs, you haveto find the output of the code for the inputs. The code is as follows:

int a, b, c, d, e, f;  
int fn( int n ) \{   
    if( n == 0 ) return a;  
    if( n == 1 ) return b;  
    if( n == 2 ) return c;  
    if( n == 3 ) return d;  
    if( n == 4 ) return e;  
    if( n == 5 ) return f;  
    return( fn(n\-1) \+ fn(n\-2) \+ fn(n\-3) \+ fn(n\-4) \+ fn(n\-5) \+ fn(n\-6) );  
\}  
int main() \{   
    int n, caseno = 0, cases;  
    scanf("%d", &cases);  
    while( cases\-- ) \{   
        scanf("%d %d %d %d%d %d %d", &a, &b, &c, &d, &e, &f, &n);  
        printf("Case %d: %d**\\n**", \++caseno, fn(n) % 10000007);  
    \}  
    return 0;  
\}

# Input #

Input starts with an integer **T (≤ 100)**,denoting the number of test cases.

Each case contains seven integers, **a, b, c, d, e, f and n**.All integers will be non-negative and **0 ≤ n ≤ 10000** and the eachof the others will be fit into a 32-bit integer.

# Output #

For each case, print the output of the given code. The given code may have integer overflow problem in the compiler, so be careful.

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   <td> <h1>Sample Input</h1> </td> 
   <td> <h1>Output for Sample Input</h1> </td> 
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   <td> <p>5</p> <p>0 1 2 3 4 5 20</p> <p>3 2 1 5 0 1 9</p> <p>4 12 9 4 5 6 15</p> <p>9 8 7 6 5 4 3</p> <p>3 4 3 2 54 5 4</p> </td> 
   <td> <p>Case 1: 216339</p> <p>Case 2: 79</p> <p>Case 3: 16636</p> <p>Case 4: 6</p> <p>Case 5: 54</p> </td> 
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</table>

题意: 让你求上面代码的运行结果.

分析: 上面的代码是递归形式, 化成递推就是 f(n) = f(n-1)+f(n-2)+f(n-3)+f(n-4)+f(n-5)+f(n-6); 那么就可以进行递推了, 但是递推还是有点慢, 因为T比较大, 递推都是可以用矩阵运算来加速的, 构造一个6乘6的矩阵, 然后快速幂运算, 把复杂度降到6\*6\*6\*log\*(n); 注意我的写法是不能处理0的情况的, 这次吃了大亏了.

#include<bitset>
    #include<map>
    #include<vector>
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<string>
    #include<algorithm>
    #include<cmath>
    #include<stack>
    #include<queue>
    #include<set>
    #define inf 0x3f3f3f3f
    #define mem(a,x) memset(a,x,sizeof(a))
    #define F first
    #define S second
    using namespace std;
    
    typedef long long ll;
    typedef unsigned long long ull;
    typedef pair<int,int> pii;
    
    inline int in()
    {
        int res=0;char c;int f=1;
        while((c=getchar())<'0' || c>'9')if(c=='-')f=-1;
        while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
        return res*f;
    }
    const int N=100010,MOD=10000007;
    int q[7],n;
    struct M
    {
        int a[6][6];
        void init()
        {
            mem(a,0);
            int j=1;
            for(int i=0;i<5;i++)
            {
                a[i][j++]=1;
            }
            for(int i=0;i<6;i++) a[5][i]=1;
        }
    };
    
    M mul(M x,M y)
    {
        M ret;
        mem(ret.a,0);
        for(int i=0;i<6;i++)
        {
            for(int k=0;k<6;k++)
            {
                if(x.a[i][k]==0) continue;
                for(int j=0;j<6;j++)
                {
                    ret.a[i][j] = (ll(1LL*x.a[i][k]*y.a[k][j] + ret.a[i][j]))%MOD;
                }
            }
        }
        return ret;
    }
    
    M solve()
    {
        M ans,x;
        ans.init();
        x.init();
        n--;
        while(n>0)
        {
            if(n & 1) ans = mul(ans,x);
            x=mul(x,x);
            n>>=1;
        }
        return ans;
    }
    
    
    int main()
    {
        int T=in(),ca=1;
        while(T--)
        {
            for(int i=0;i<6;i++) q[i]=in()%MOD;
            n=in();
            if(n==0) 特判!!
            {
                printf("Case %d: %d\n", ca++, q[0]);
                continue;
            }
            M ans=solve();
    
            int ret=0;
            for(int i=0;i<6;i++) ret = ((ll)(ret+1LL*ans.a[0][i]*q[i]))%MOD;
    
            printf("Case %d: %d\n", ca++, ret);
        }
        return 0;
    }