题目：http://www.luogu.org/problem/show?pid=2265
题解：ans=C(n+m，n）%p
求一下逆元就行
代码；
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<cmath> 5 #include<iostream> 6 #include<string> 7 #include<set> 8 #include<map> 9 #include<vector> 10 #include<algorithm> 11 #include<queue> 12 #define for0(i,n) for(int i=0;i<=n;i++) 13 #define for1(i,n) for(int i=1;i<=n;i++) 14 #define for2(i,x,y) for(int i=(x);i<=(y);i++) 15 #define for3(i,y,x) for(int i=(y);i>=(x);i--) 16 #define maxn 100000+5 17 #define mod 1000000007 18 #define ll long long 19 using namespace std; 20 inline int read() 21 { 22 int x=0,f=1;char ch=getchar(); 23 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 24 while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();} 25 return x*f; 26 } 27 ll power(ll x,ll y) 28 { 29 ll t=1; 30 for(;y;y>>=1,x=(x*x)%mod) 31 if(y&1)t=(t*x)%mod; 32 return t; 33 } 34 int main() 35 { 36 freopen("input.txt","r",stdin); 37 freopen("output.txt","w",stdout); 38 ll n=read(),m=read(); 39 ll t1=1,t2=1; 40 for1(i,n)t1=(t1*(ll)i)%mod; 41 for2(i,m+1,n+m)t2=(t2*(ll)i)%mod; 42 printf("%lld ",t2*power(t1,mod-2)%mod); 43 return 0; 44 }