先丢着。以后再来填坑。
${{n+1}\choose{n+m+1}}-2\times {{n+1}\choose{n+m}}$
$={{n}\choose{n+m}}+{{n+1}\choose{n+m}}-2\times {{n+1}\choose{n+m}}$
$={{n}\choose{n+m}}-{{n+1}\choose{n+m}}$
$=\frac{(n+m)!}{n!m!} - \frac{(n+m)!}{(n+1)!(m-1)!}$
$=\frac{(n+m)!}{n!m!} - \frac{m}{n+1}\times\frac{(n+m)!}{n!m!}$
$=\frac{n-m+1}{n+1}\times\frac{(n+m)!}{n!m!}$
```cpp
#include <cstdio>
#define mod 20100403
#define Lovelive long long
inline Lovelive pow(Lovelive x, int p = mod - 2) {
Lovelive ret = 1;
for(; p; p >>= 1, x = x * x % mod) if(p & 1) ret = ret * x % mod;
return ret;
}
int main() {
Lovelive ans, fac;
register int n, m, i;
scanf("%d%d", &n, &m);
for(i = ans = fac = 1; i <= m; ++i) fac = fac * i % mod;
ans = pow(fac);
for(; i <= n; ++i) fac = fac * i % mod;
ans = ans * pow(fac) % mod;
for(; i <= m + n; ++i) fac = fac * i % mod;
ans = ans * fac % mod * (n - m + 1) % mod * pow(n + 1) % mod;
printf("%d", (int)ans);
}
```