题解：http://www.cnblogs.com/jianglangcaijin/p/3443689.html
求逆元还是快速幂好用
1 #include<algorithm> 2 #include<iostream> 3 #include<cstdlib> 4 #include<cstring> 5 #include<cstdio> 6 #include<string> 7 #include<cmath> 8 #include<ctime> 9 #include<queue> 10 #include<stack> 11 #include<map> 12 #include<set> 13 #define rre(i,r,l) for(int i=(r);i>=(l);i--) 14 #define re(i,l,r) for(int i=(l);i<=(r);i++) 15 #define Clear(a,b) memset(a,b,sizeof(a)) 16 #define inout(x) printf("%d",(x)) 17 #define douin(x) scanf("%lf",&x) 18 #define strin(x) scanf("%s",(x)) 19 #define LLin(x) scanf("%lld",&x) 20 #define op operator 21 #define CSC main 22 typedef unsigned long long ULL; 23 typedef const int cint; 24 typedef long long LL; 25 using namespace std; 26 void inin(int &ret) 27 { 28 ret=0;int f=0;char ch=getchar(); 29 while(ch<'0'||ch>'9'){if(ch=='-')f=1;ch=getchar();} 30 while(ch>='0'&&ch<='9')ret*=10,ret+=ch-'0',ch=getchar(); 31 ret=f?-ret:ret; 32 } 33 int n,m; 34 const LL mod=20100403; 35 LL inv(LL a,LL n) 36 { 37 a%=n; 38 LL b=n-2,ret=1; 39 while(b) 40 { 41 if(b&1)ret=ret*a%n; 42 b>>=1; 43 a=a*a%n; 44 } 45 return ret; 46 } 47 LL C(int a,int b) 48 { 49 LL x=a; 50 re(i,2,b)x=x*(a-i+1)%mod*inv(i,mod)%mod; 51 return x; 52 } 53 int CSC() 54 { 55 inin(n),inin(m); 56 printf("%lld",((C(n+m,n)-C(n+m,m-1))%mod+mod)%mod); 57 return 0; 58 }