HDU 5793 - A Boring Question

HDU 5793 - A Boring Question
题意：

　　计算 ( ∑(0≤K₁,K₂...K_m≤n )∏(1≤j<m) C[K_j, K_j+1]  ) % 1000000007=? (C[Kj, Kj+1] 为组合数)
    
分析:

　　利用二项式展开： (a + b) ^ n =  ∑(r = 0, n) (C[n, r] * a^(n-r) * b^r )
    
    化简：
           ∑(0≤K₁,K₂...K_m≤n )∏(1≤j<m) C[K_j, K_j+1]
    
         = ∑( K_m = 0, n ) ∑( K_m-1 = 0, Km ) ∑( K_m-2 = 0, K_m-1 )...∑( K₁ = 0, K₂ ) ( C[K_m, K_m-1] * C[K_m-1, K_m-2] *...*C[K₂, K₁]  )
           
         = ∑( K_m = 0, n ) ∑( K_m-1 = 0, Km )C[K_m, K_m-1]  ∑( K_m-2 = 0, K_m-1 )C[K_m-1, K_m-2] ... ∑( K₂ = 0, K₃ )C[K₃, K₂]  ∑( K₁ = 0, K₂)C[K₂, K₁]   //后面的积可分别提到和式前面
                                                                                                                                    
         = ∑( K_m = 0, n ) ∑( K_m-1 = 0, Km )C[K_m, K_m-1]  ∑( K_m-2 = 0, K_m-1 )C[K_m-1, K_m-2] ... ∑( K₂ = 0, K₃ ) ( C[K₃, K₂] * 2^K₂ )

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　 // ∑( K₁ = 0, K₂)C[K₂, K₁]  为(1 + 1) ^ k₂ 的二项式展开
                                                                                                                                    
         = ∑( K_m = 0, n ) m ^ K_m          //∑( K₂ = 0, K₃ ) ( C[K₃, K₂] * 2^K₂ ) 为 (1 + 2) ^ k₃ 的二项式展开 ，接下来依次向上化简
         
         = ( m^(n+1) - 1 ) / ( m - 1 ) 　　//等比数列求和公式
         
    接下来求快速幂和逆元即可.

　　（博客园怎么连个公式编辑器都没有= =）

 #include <iostream>
 #include <cstdio>
 using namespace std;
 #define LL long long
 const LL MOD = ;
 LL PowMod(LL a, LL p, LL MOD)
 {
     int res = ;
     while(p)
     {
         if(p&) res = (res * a) %MOD;
         p >>= ;
         a = (a * a) % MOD;
     }
     return res;
 }
 int t;
 LL n,m;
 int main()
 {
     scanf("%d",&t);
     while(t--)
     {
         scanf("%lld%lld", &n, &m);
         LL ans = PowMod(m, n+, MOD);
         --ans;
         LL inv = PowMod(m-, MOD-, MOD);
         ans = (ans * inv) % MOD;
         printf("%lld\n",ans);
     }
 }