HDU 2940 Hex factorial (simple high precision)

Test instructions: Ask for a n! The number of 16 binary representations of 0.

High-precision simulations, or direct java. Here you can give a high-precision template.

#include <cstdio> #include <cstdio> #include <cmath> #include <queue> #include <stack># include<string> #include <cstring> #include <iostream> #include <map> #include <vector># include<algorithm> #include <set> #include <cmath>using namespace std;const int mmax = 10010;const int    INF = 0x3fffffff;struct bignum{int Sz;    int Num[mmax];        void print () {for (int i=sz-1;i>=0;i--) printf ("%0x", Num[i]);    Puts ("");        } bignum () {} bignum (int Sz,char *a) {sz=sz;        for (int i=0;i<sz;i++) {num[i]=a[sz-1-i]-' 0 ';        }} bignum (int sz,int *a) {sz=sz;    for (int i=0;i<sz;i++) num[i]=a[i];        } bignum (int x) {sz=0;            while (x) {num[sz++]=x%16;        x/=16;        }} Bignum operator + (const bignum &a) {int Tmp[mmax];        memset (tmp,0,sizeof tmp); int Len=max(SZ,A.SZ);            for (int i=0;i<len;i++) {tmp[i]+=num[i]+a.num[i];            TMP[I+1]+=TMP[I]/10;        tmp[i]%=10;    } return Bignum (len+ (tmp[len]?1:0), TMP);        } bignum Operator-(const bignum &a) {int Tmp[mmax];        memset (tmp,0,sizeof tmp);        int Len=max (SZ,A.SZ);            for (int i=0;i<len;i++) {tmp[i]+=num[i]-a.num[i];                if (tmp[i]<0) {tmp[i+1]-=1;            tmp[i]+=10;        }} for (int i=len-1;i>=0;i--) {if (Tmp[i]) return Bignum (I+1,TMP);    } return Bignum (1,TMP);        } bignum operator * (const bignum &a) {int Tmp[mmax];        memset (tmp,0,sizeof tmp);            for (int i=0;i<sz;i++) for (int j=0;j<a.sz;j++) {tmp[i+j]+=num[i]*a.num[j];  } for (int i=0;i<sz+a.sz;i++) {tmp[i+1]+=tmp[i]/16;          tmp[i]%=16;        } for (int i=sz+a.sz-1;i>=0;i--) {if (Tmp[i]) return Bignum (I+1,TMP);    } return Bignum (1,TMP);    }};int Main () {int n;        while (Cin>>n && n>=0) {bignum ans (1);        for (int i=1;i<=n;i++) ans=ans*bignum (i);        int cnt=0; for (int I=0;i<ans.        sz;i++) {if (ans.num[i]==0) cnt++;    } printf ("%d\n", CNT); } return 0;}

HDU 2940 Hex factorial (simple high precision)