hdu3658 How many words

How many words

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 259    Accepted Submission(s): 128

Problem Description

In order to make a new word, we will pick out m letters from all the upper case letters and lower case letters(from `a' to `Z'). Therefore, that means you can pick some same letters. But here are two rules:
● as to all the neighbour letters, the absolute value of their ASCII code must be not greater than 32.
● there must be at least one pair of neighbour letters whose absolute value of ASCII code is exactly equal to 32. For example, considering the word in the form like "Xx" or "xX", the neighbour letters have an absolute value of ASCII code exactly equal to 32.
Now how many di erent words can we get?

 

Input

The first line of input is the number of test case. For each test case, there is only one line contains one integer m(2 ≤ m ≤ 10⁹).

 

Output

For each test case output one line, just the answer mod 1000000007.

 

Sample Input

4231007926778

 

Sample Output

524056533550434773908369

 

Author

windy7926778

 

Source

2010 Asia Regional Chengdu Site —— Online Contest

 

Recommend

lcy

简单的矩阵快速幂。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define MOD 1000000007typedef struct{    long long a[55][55];}Matrix;Matrix Mul(Matrix x,Matrix y){    int i,j,k;    Matrix tag;    for (i=0;i<52;i++)    {        for (j=0;j<52;j++)        {            tag.a[i][j]=0;            for (k=0;k<52;k++)            {                tag.a[i][j]+=x.a[i][k]*y.a[k][j];                tag.a[i][j]%=MOD;            }        }    }    return tag;}Matrix a,b;int main(){    int i,j,n,T,p;    long long ans;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        for (i=0;i<52;i++)        {            for (j=0;j<52;j++)            {                a.a[i][j]=(abs(i-j)<=26);                b.a[i][j]=0;            }            b.a[i][0]=1;        }        p=n;        n--;        while(n)        {            if (n&1) b=Mul(a,b);            a=Mul(a,a);            n>>=1;        }        ans=0;        for (i=0;i<52;i++)        {            ans+=b.a[i][0];            ans%=MOD;        }        for (i=0;i<52;i++)        {            for (j=0;j<52;j++)            {                a.a[i][j]=(abs(i-j)<26);                b.a[i][j]=0;            }            b.a[i][0]=1;        }        n=p;        n--;        while(n)        {            if (n&1) b=Mul(a,b);            a=Mul(a,a);            n>>=1;        }        for (i=0;i<52;i++)        {            ans+=MOD-b.a[i][0];            ans%=MOD;        }        printf("%I64d\n",ans);    }    return 0;}