题面
题解
任意两个障碍不在同一列
要求你放$N$个棋子也满足每行只有一枚棋子,每列只有一枚棋子的限制。
这™不就是个错排吗???
$$ h_i=(n-1)(h_{i-1}+h_{i-2}),h_1=0,h_2=1 $$
写个高精度就好了。。。
代码
#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#define RG register
inline int read()
{
int data = 0, w = 1; char ch = getchar();
while(ch != '-' && (!isdigit(ch))) ch = getchar();
if(ch == '-') w = -1, ch = getchar();
while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
return data * w;
}
const int maxn(210), Mod(1e8);
int n, a[maxn];
long long f[maxn][50];
int main()
{
n = read(); f[1][0] = 0, f[2][0] = 1;
for(RG int i = 3; i <= n; i++)
{
for(RG int j = 0; j <= a[i - 1]; j++)
f[i][j] = f[i - 1][j] + f[i - 2][j];
a[i] = a[i - 1];
for(RG int j = 0; j <= a[i - 1]; j++)
f[i][j + 1] += f[i][j] / Mod,
f[i][j] %= Mod;
while(f[i][a[i] + 1]) ++a[i];
for(RG int j = 0; j <= a[i]; j++) f[i][j] *= i - 1;
for(RG int j = 0; j <= a[i]; j++)
f[i][j + 1] += f[i][j] / Mod,
f[i][j] %= Mod;
while(f[i][a[i] + 1]) ++a[i];
}
printf("%lld", f[n][a[n]]);
for(RG int i = a[n] - 1; ~i; i--) printf("%08lld", f[n][i]);
return 0;
}