luoguP2265 路边的水沟

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；

 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<cmath>
 #include<iostream>
 #include<string>
 #include<set>
 #include<map>
 #include<vector>
 #include<algorithm>
 #include<queue>
 #define for0(i,n) for(int i=0;i<=n;i++)
 #define for1(i,n) for(int i=1;i<=n;i++)
 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 #define for3(i,y,x) for(int i=(y);i>=(x);i--)
 #define maxn 100000+5
 #define mod 1000000007
 #define ll long long
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}
     return x*f;
 }
 ll power(ll x,ll y)
 {
     ll t=;
     for(;y;y>>=,x=(x*x)%mod)
      if(y&)t=(t*x)%mod;
     return t;
 }
 int main()
 {
     freopen("input.txt","r",stdin);
     freopen("output.txt","w",stdout);
     ll n=read(),m=read();
     ll t1=,t2=;
     for1(i,n)t1=(t1*(ll)i)%mod;
     for2(i,m+,n+m)t2=(t2*(ll)i)%mod;
     printf("%lld\n",t2*power(t1,mod-)%mod);
     return ;
 }