hoj3152-Dice 等比数列求和取模

http://acm.hit.edu.cn/hoj/problem/view?id=3152

Dice
My Tags      (Edit)
    Source :
    Time limit :  sec         Memory limit :  M

Submitted : , Accepted : 

    You have a dice with M faces, each face contains a distinct number.

    Your task is to calculate the expected number of tosses until a number facing up for consecutive N times.

    We assume when we tossing the dice, each face will occur randomly and uniformly.
Input

  Each test cases contains only one line with two integers M, N. (<=M, N<=)
Output

   For each test case, display a single line with the answer to the question above, the answer maybe very large, you need to MOD it with  .

Sample Input

Sample Output

题目

题目说是求M面骰子投出N次相同点数所需要投的次数的期望值，看样例可以知道其实是求1+M+M^2+.......+M^(N-1)，等比数列求和取模！

这个好像是某次校赛的题，当时我是直接等比数列慢慢加起来，没用等比数列求和公式，就得了。不过现在这题数据好像加强了…不能这样撸啦！而等比数列求和公式又有除法，不好搞取模。于是我也不会了。

然后我找到了一个叫kk303的人写的这个题解：

http://blog.csdn.net/kk303/article/details/9332513

虽然还是不太懂为什么不过得到了超碉的公式：

求等比为k的等比数列之和T[n]..当n为偶数..T[n] = T[n/2] + pow(k,n/2) * T[n/2]

n为奇数...T[n] = T[n/2] + pow(k,n/2) * T[n/2] + 等比数列第n个数的值

比如 1+2+4+8 = (1+2) + 4*(1+2)

1+2+4+8+16 = (1+2) + 4*(1+2) + 16

哇，终于A了，简直屁滚尿流

代码：

 #include<stdio.h>
 #define MO 1000000007
 long long m,n;
 long long powmod(long long a,long long b)
 {
     long long c=;
     while(b>)
     {
         if(b%!=)
             c=c*a%MO;
         a=a*a%MO;
         b=b/;
     }
     return c;
 }

 long long T(long long n)
 {
     if(n<=) return ;
     long long TN2=T(n/);
     if(n%==)
     {
         return (TN2 + powmod(m,n/) * TN2)%MO;
     }
     else
     {
         return (TN2 + powmod(m,n/) * TN2 + powmod(m,n-))%MO;
     }
 }

 int main()
 {
     long long ans,k;
     long long i,j,t;
     scanf("%lld",&t);
     for(i=; i<=t; i++)
     {
         scanf("%lld%lld",&m,&n);
         if(m==) ans=n;
         else
         {
             ans=T(n);
         }
         printf("%lld\n",ans);
     }
     return ;
 }