rt，求解```cpp #include #include #include #include #include #include #include #include #include #define f(c,a,b) for(int c=a; c<=b; c++) #define N 2000010 #define pii pair<int,int> #define mp make_pair #define fi first #define se second #define sl (u<<1) #define sr ((u<<1)+1) #define pb push_back #define eps 1e-8

using namespace std; typedef double db; typedef long long ll; const int mo = 20100403; ll jc[N]={1};

ll fp(ll a, ll b){ // a^b ll t = a, ans = 1; for (int i=0; (1LL<<i)<=b; i++) { if( (1LL<<i)&b ) ans = anst % mo; t = tt % mo; } return ans; } //费马小定理 ll my_inv(ll a) { return fp(a,mo-2); } //扩展欧几里得 ll c(ll n, ll m) { return jc[n]my_inv(jc[m])%momy_inv(jc[n-m])%mo; }

int main(){ // freopen("owo.in","r",stdin); // ios::sync_with_stdio(false); // cin.tie(nullptr); int n,m; cin>>n>>m; if(m==1){ cout<<1<<endl; return 0; } f(i,1,N-1) jc[i]=(jc[i-1]*i)%mo; ll a=c(n+m,n), b=c(n+m,n+1); // cout<<a<<' '<<b<<endl; cout<< ( (a-b)%mo+mo )%mo << endl; return 0; }