hdu 6030 Happy Necklace

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

223

 

Sample Output

34

题意：有红蓝两种颜色的珠子，要组成一个长串（不成环），对任意连续素数数个串需满足红珠子数不大于蓝珠子数，给出串的长度统计有多少种情况。

思路：推出递推式f(n) = f(n-1)+f(n-3)，然后构造矩阵，矩阵快速幂

#include<stdio.h>#include<string.h>#include<math.h>#include<string.h>#include<stdlib.h>#include<algorithm>#include<iostream>#include<queue>#include<stack>using namespace std;const double EXP = 1e-9;typedef long long ll;const ll maxn =  3;  //确定矩阵大小const ll mod =   1e9+7;  //取余数用struct mat{    ll m[maxn][maxn];}unit;   //unit为单位矩阵mat matmul(mat a,mat b)  //俩矩阵相乘{    mat ans;    ll x,i,j,k;    for(i=0;i<maxn;i++)    {        for(j=0;j<maxn;j++)        {            x=0;            for(k=0;k<maxn;k++)            {                x+=(a.m[i][k]*b.m[k][j])%mod;            }            ans.m[i][j] = x%mod;        }    }    return ans;}void init_unit()   //初始化单位矩阵{    for(ll i=0;i<maxn;i++)        unit.m[i][i] = 1;    return ;}mat pow_mat(mat a,ll n)  //矩阵快速幂，求矩阵a的n次幂{    mat ans = unit;  //unit为幺元    while(n)    {        if(n&1)            ans=matmul(ans,a);        a = matmul(a,a);        n>>=1;    }    return ans;}int main(){    ll n,t;    init_unit();    scanf("%lld",&t);    while(t--)    {        scanf("%lld",&n);        if(n==2)            printf("3\n");        else if(n==3)            printf("4\n");        else if(n==4)            printf("6\n");        else        {            mat a,b;   //a为表达式矩阵，b为快速幂矩阵            b.m[0][0]=1,b.m[0][1]=1,b.m[0][2]=0;            b.m[1][0]=0,b.m[1][1]=0,b.m[1][2]=1;            b.m[2][0]=1,b.m[2][1]=0,b.m[2][2]=0;            a.m[0][0]=6, a.m[0][1]=4,a.m[0][2]=3;            b = pow_mat(b,n-4);            a = matmul(a,b);            printf("%lld\n",a.m[0][0]%mod);        }    }    return 0;}

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

223

 

Sample Output

34