首先很容易想到暴力DP
设状态f[i][j]表示当前放了第i个数，最大的数为j的方案数。
然后根据转移推出实际上是在下图走路的方案数
[left( left( egin{matrix}
x + y - 2 \ x - 1
end{matrix}
ight)-
left( egin{matrix}
x+y-2 \ x-2
end{matrix}
ight)
ight) imes
left(left( egin{matrix}
2n-x-y \ n-x
end{matrix}
ight)-
left( egin{matrix}
2n-x-y \ n-x+1
end{matrix}
ight)
ight)
]
注意：
以上情况是 (x leq y) 的情况，对于(x > y)的情况，显然需要处理，
由于该图的对称性，我们珂以轻易地
将 (x) 变换为 (n - x - 1)；
将 (y) 变换为 (n - y - 1)。
#include <cstdio>
#define MOD 1000000007
namespace fast_IO{
const int IN_LEN = 10000000, OUT_LEN = 10000000;
char ibuf[IN_LEN], obuf[OUT_LEN], *ih = ibuf + IN_LEN, *oh = obuf, *lastin = ibuf + IN_LEN, *lastout = obuf + OUT_LEN - 1;
inline char getchar_(){return (ih == lastin) && (lastin = (ih = ibuf) + fread(ibuf, 1, IN_LEN, stdin), ih == lastin) ? EOF : *ih++;}
inline void putchar_(const char x){if(oh == lastout) fwrite(obuf, 1, oh - obuf, stdout), oh = obuf; *oh ++= x;}
inline void flush(){fwrite(obuf, 1, oh - obuf, stdout);}
int read(){
int x = 0; char ch = ' ';
while (ch < '0' || ch > '9') ch = getchar_();
while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar_(); return x;
}
void write(int x){
if (x < 0) putchar_('-'), x = -x;
if (x > 9) write(x / 10);
putchar_(x % 10 + '0');
}
}
using namespace fast_IO;
long long fc[20000005];
long long ksm(long long a, long long b){
long long res = 1;
for ( ; b; b >>= 1, a = (a * a) % MOD)
if (b & 1) res = (res * a) % MOD;
return res;
}
long long getC(int a, int b){
return fc[b] * ksm(fc[b - a], MOD - 2) % MOD * ksm(fc[a], MOD - 2) % MOD;
}
int main(){
fc[0] = fc[1] = 1;
for (register int i = 2; i <= 20000000; ++i)
fc[i] = fc[i - 1] * i % MOD;
int T = read();
while (T--){
long long n = read(), x = read(), y = read();
if (x > y){
x = n - x + 1;
y = n - y + 1;
}
long long res = (getC(x - 1, x + y - 2) - getC(x - 2, x + y - 2))
* (getC(n - x, 2 * n - x - y) - getC(n - x + 1, 2 * n - x - y)) % MOD;
(res < 0) && (res += MOD);
write(res), putchar_('
');
}
flush(); return 0;
}