HDU-6030-Happy Necklace

本是古典何须时尚 2022-05-17 01:45 99阅读 0赞

# Happy Necklace #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)  
Total Submission(s): 1693    Accepted Submission(s): 715**

**Problem Description**

Little Q wants to buy a necklace for his girlfriend. Necklaces are single strings composed of multiple red and blue beads.  
Little Q desperately wants to impress his girlfriend, he knows that she will like the necklace only if for every prime length continuous subsequence in the necklace, the number of red beads is not less than the number of blue beads.  
Now Little Q wants to buy a necklace with exactly n beads. He wants to know the number of different necklaces that can make his girlfriend happy. Please write a program to help Little Q. Since the answer may be very large, please print the answer modulo 109+7.  
Note: The necklace is a single string, **\{not a circle\}**.

**Input**

The first line of the input contains an integer T(1≤T≤10000), denoting the number of test cases.  
For each test case, there is a single line containing an integer n(2≤n≤1018), denoting the number of beads on the necklace.

**Output**

For each test case, print a single line containing a single integer, denoting the answer modulo 109+7.

**Sample Input**

2 2 3

**Sample Output**

3 4

题目大意：一个手链只有红色和蓝色的珠子，假设红色是1，蓝色是0，在任意的素数个数的珠子中红色大于等于蓝色，分析可知若要满足条件只需要考虑长度为2和长度为3的串。

所以我们可以得到推导公式：f\[n\]=f\[n-1\]+f\[n-3\];

由于n的取值范围比较大，所以我们可以采用矩阵快速幂来做

推导出初始矩阵为：

![70][]然后全部加起来就可以了

所以代码：

#include<map>
    #include<stack>
    #include<cmath>
    #include<queue>
    #include<string>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    #define ll long long
    const int maxn=1010;
    const int inf=0x3f3f3f;
    const int mod=1e9+7;
    const int N=3;
    ll res[N][N];
    void juzhen(ll a[N][N],ll b[N][N])
    {
        ll c[N][N],i,j,k;
        memset(c,0,sizeof(c));
        for(k=0; k<N; k++)
        {
            for(i=0; i<N; i++)
            {
                for(j=0; j<N; j++)
                {
                    c[k][i]+=a[k][j]*b[j][i];
                    c[k][i]%=mod;
                }
            }
        }
        for(i=0; i<N; i++)
        {
            for(j=0; j<N; j++)
                a[i][j]=c[i][j];
        }
    }
    void ppow(ll a[N][N],ll n)
    {
        memset(res,0,sizeof(res));
        for(int i=0; i<N; i++)
            res[i][i]=1;
        while(n)
        {
            if(n&1)
                juzhen(res,a);
            juzhen(a,a);
            n>>=1;
        }
        ll x = (res[0][0]+res[0][1]+res[0][2]) % mod;
        ll y = (res[1][0]+res[1][1]+res[1][2]) % mod;
        ll z = (res[2][0]+res[2][1]+res[2][2]) % mod;
        printf("%lld\n",(x + y + z)%mod);
    }
    int main()
    {
        int test;
        scanf("%d",&test);
        while(test--)
        {
            ll n;
            scanf("%lld",&n);
            if(n==2)
            {
                printf("3\n");
                continue;
            }
            ll a[N][N];
            a[0][0] = 1; a[0][1] = 0; a[0][2] = 1;
    		a[1][0] = 1; a[1][1] = 0; a[1][2] = 0;
    		a[2][0] = 0; a[2][1] = 1; a[2][2] = 0;
            ppow(a,n-2);
        }
        return 0;
    }

[70]: /images/20220517/ef13193ad1eb4264bd3336f3a59e7c75.png