Bzoj 2024 SHOI2009 Prom Dynamic planning + principle + high precision

Topic: Given two sequence, how many matches satisfy a[i]<b[i] The logarithm does not exceed K pair

See http://blog.csdn.net/popoqqq/article/details/44514113

High precision has been wasted ...

#include <cstdio> #include <cstring> #include <iomanip> #include <iostream> #include < Algorithm> #define M 202using namespace std;struct big_int{#define BASE 1000int num[125],cnt; Big_int () {}big_int (Int _) {memset (num,0,sizeof num); num[cnt=1]=_;} big_int& operator = (Int x) {return * (new (this) big_int (x));} big_int& operator + = (const big_int &x) {Int I;cnt=max (cnt,x.cnt); for (i=1;i<=cnt;i++) {num[i]+=x.num[i];if ( num[i]>=base) Num[i+1]+=num[i]/base,num[i]%=base;} if (num[cnt+1]) ++cnt;return *this;} Big_int operator + (const big_int &x) Const{big_int Re=*this;re+=x;return re;} big_int& operator-= (const big_int &x) {Int i;for (i=1;i<=cnt;i++) {num[i]-=x.num[i];while (num[i]<0) num[ I+1]--, num[i]+=base;} while (cnt>0 &&!num[cnt])--cnt;return *this;} Big_int operator-(const big_int &x) Const{big_int Re=*this;re+=x;return re;} Friend Big_int operator * (const big_int &x,const big_int &y) {Int i,j; Big_int Re=0;for (i=1;i< =x.cnt;i++) for (j=1;j<=y.cnt;j++) re.num[i+j-1]+=x.num[i]*y.num[j];re.cnt=x.cnt+y.cnt;for (i=1;i<=re.cnt;i + +) if (re.num[i]>=base) re.num[i+1]+=re.num[i]/base,re.num[i]%=base;if (!re.num[re.cnt]) Re.cnt--;return re;} big_int& operator *= (const big_int &x) {return *this=*this*x;} Friend ostream& operator << (ostream &_,big_int &x) {Int i;printf ("%d", x.num[x.cnt]); for (i=x.cnt-1;i >0;i--) printf ("%03d", X.num[i]); return _;}} F[m][m],g[m],c[m][m],fac[m],ans;int n,k;int A[m],b[m],next[m];int Main () {int i,j;cin>>n>>k;for (i=1;i <=n;i++) scanf ("%d", &a[i]), for (i=1;i<=n;i++) scanf ("%d", &b[i]); sort (a+1,a+n+1); sort (b+1,b+n+1); (j=1,i=1;i<=n;i++) {for (; j<=n&&b[j]<=a[i];j++); next[i]=j-1;} for (i=0;i<=n;i++) {c[i][0]=1;for (j=1;j<=i;j++) c[i][j]=c[i-1][j]+c[i-1][j-1];} for (fac[0]=1,i=1;i<=n;i++) fac[i]=fac[i-1]*i;f[0][0]=1;for (i=1;i<=n;i++) {f[i][0]=1;for (j=1;j<=i;j++) f[ I][j]=f[i-1][j]+f[i-1][j-1]*max (next[i]-(j-1), 0);} for (i=n;i>=n-k;i--) {g[i]=f[n][i]*fac[n-i];for (j=i+1;j<=n;j++) g[i]-=g[j]*c[j][i];ans+=g[i];} Cout<<ans<<endl;return 0;}

Bzoj 2024 SHOI2009 Prom Dynamic planning + principle + high precision