BZOJ 2024: [SHOI2009] 舞会 [容斥原理 高精度]
题意:和上题基本一样,求至少k对a>b的方案数。不取模!!!
做k+1遍容斥就行了
高精度超强!!!几乎把所有的都用上了
然后,注意有负数,所以容斥的时候正负分别保存然后再一减就行了
~~这是我省选前最后一次写高精度了~~
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N=205, B=1e4;
inline int read(){
char c=getchar();int x=0,f=1;
while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
return x*f;
}
struct Big{
int a[120], n;
int& operator [](int x) {return a[x];}
Big():n(1) {memset(a, 0, sizeof(a));}
void ini(int x) {a[1]=x; n=1;}
}f[N][N], C[N][N], fac[N];
Big operator *(Big a, int b) {
int g=0;
for(int i=1; i<=a.n; i++)
g += a[i]*b, a[i] = g%B, g/=B;
if(g) a[++a.n] = g;
return a;
}
Big operator *(Big a, Big b) {
Big c;
for(int i=1; i<=a.n; i++) {
int g=0;
for(int j=1; j<=b.n; j++)
g += c[i+j-1]+a[i]*b[j], c[i+j-1] = g%B, g/=B;
c[i+b.n] = g;
}
c.n = a.n + b.n;
while(c.n>1 && c[c.n]==0) c.n--;
return c;
}
Big operator +(Big a, Big b) {
int g=0, n=max(a.n, b.n);
for(int i=1; i<=n; i++) {
g += i<=a.n ? a[i] : 0;
g += i<=b.n ? b[i] : 0;
a[i] = g%B, g/=B;
}
a.n = n;
if(g) a[++a.n] = g;
return a;
}
Big operator -(Big a, Big b) {
for(int i=1; i<=b.n; i++) {
if(a[i]<b[i]) a[i]+=B, a[i+1]--;
a[i] -= b[i];
}
int p=b.n+1;
while(a[p]<0) a[p]+=B, a[++p]--;
while(a.n>1 && a[a.n]==0) a.n--;
return a;
}
void Print(Big &a) {
printf("%d", a[a.n]);
for(int i=a.n-1; i>=1; i--) printf("%04d", a[i]);
}
int n, k, a[N], b[N];
int g[N];
void dp() {
fac[0].ini(1);
for(int i=1; i<=n; i++) fac[i] = fac[i-1]*i;
C[0][0].ini(1);
for(int i=1; i<=n; i++) {
C[i][0].ini(1);
for(int j=1; j<=n; j++) C[i][j] = C[i-1][j] + C[i-1][j-1];
}
int now=0;
for(int i=1; i<=n; i++) {
while(now<n && a[i]>b[now+1]) now++;
g[i] = now;
}
for(int i=0; i<=n; i++) f[i][0].ini(1);
for(int i=1; i<=n; i++)
for(int j=1; j<=g[i]; j++) f[i][j] = f[i-1][j] + f[i-1][j-1]*(g[i]-j+1);// Print(f[i][j]), puts("");
}
int main() {
freopen("in","r",stdin);
n=read(); k=read();
for(int i=1; i<=n; i++) b[i]=read();
for(int i=1; i<=n; i++) a[i]=read();
sort(a+1, a+1+n); sort(b+1, b+1+n);
dp();
Big pos, neg, ans;
int kkk=k;
for(int k=0; k<=kkk; k++) { //printf("kkkkkk %d\n",k);
for(int i=k; i<=n; i++){ //printf("i %d\n",i);
Big t = fac[n-i] * f[n][i] * C[i][k]; //printf("t ");Print(t); puts("");
if((i-k)&1) neg = neg + t;
else pos = pos + t;
}
}
ans = pos - neg;
//printf("hi %d ",i),Print(ans), puts("");
Print(ans);
}
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