题目描述
求n!中某个数码出现的次数。
输入输出格式
输入格式：
第一行为t(≤10)，表示数据组数。接下来t行，每行一个正整数n(≤1000)和数码a。
输出格式：
对于每组数据，输出一个整数，表示n!中a出现的次数。
输入输出样例
输入样例#1
2
5 2
7 0
输出样例#1
1
2
思路
重载：0分TLE
#include<string>
#include<iostream>
#include<iosfwd>
#include<cmath>
#include<cstring>
#include<stdlib.h>
#include<stdio.h>
#include<cstring>
#define MAX_L 10005 //最大长度，可以修改
using namespace std;
class bign//大整数类模版，用重载写的（因为我懒得打高精度了（逃））
{
public:
int len, s[MAX_L];//数的长度，记录数组
//构造函数
bign();
bign(const char*);
bign(int);
bool sign;//符号 1正数 0负数
string toStr() const;//转化为字符串，主要是便于输出
friend istream& operator>>(istream &,bign &);//重载输入流
friend ostream& operator<<(ostream &,bign &);//重载输出流
//重载复制
bign operator=(const char*);
bign operator=(int);
bign operator=(const string);
//重载各种比较
bool operator>(const bign &) const;
bool operator>=(const bign &) const;
bool operator<(const bign &) const;
bool operator<=(const bign &) const;
bool operator==(const bign &) const;
bool operator!=(const bign &) const;
//重载四则运算
bign operator+(const bign &) const;
bign operator++();
bign operator++(int);
bign operator+=(const bign&);
bign operator-(const bign &) const;
bign operator--();
bign operator--(int);
bign operator-=(const bign&);
bign operator*(const bign &)const;
bign operator*(const int num)const;
bign operator*=(const bign&);
bign operator/(const bign&)const;
bign operator/=(const bign&);
//四则运算的衍生运算
bign operator%(const bign&)const;//取模（余数）
bign factorial()const;//阶乘
bign Sqrt()const;//整数开根（向下取整）
bign pow(const bign&)const;//次方
//一些乱乱的函数
void clean();
~bign();
};
#define max(a,b) a>b ? a : b
#define min(a,b) a<b ? a : b
bign::bign()
{
memset(s, 0, sizeof(s));
len = 1;
sign = 1;
}
bign::bign(const char *num)
{
*this = num;
}
bign::bign(int num)
{
*this = num;
}
string bign::toStr() const
{
string res;
res = "";
for (int i = 0; i < len; i++)
res = (char)(s[i] + '0') + res;
if (res == "")
res = "0";
if (!sign&&res != "0")
res = "-" + res;
return res;
}
istream &operator>>(istream &in, bign &num)
{
string str;
in>>str;
num=str;
return in;
}
ostream &operator<<(ostream &out, bign &num)
{
out<<num.toStr();
return out;
}
bign bign::operator=(const char *num)
{
memset(s, 0, sizeof(s));
char a[MAX_L] = "";
if (num[0] != '-')
strcpy(a, num);
else
for (int i = 1; i < strlen(num); i++)
a[i - 1] = num[i];
sign = !(num[0] == '-');
len = strlen(a);
for (int i = 0; i < strlen(a); i++)
s[i] = a[len - i - 1] - 48;
return *this;
}
bign bign::operator=(int num)
{
char temp[MAX_L];
sprintf(temp, "%d", num);
*this = temp;
return *this;
}
bign bign::operator=(const string num)
{
const char *tmp;
tmp = num.c_str();
*this = tmp;
return *this;
}
bool bign::operator<(const bign &num) const
{
if (sign^num.sign)
return num.sign;
if (len != num.len)
return len < num.len;
for (int i = len - 1; i >= 0; i--)
if (s[i] != num.s[i])
return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i]));
return !sign;
}
bool bign::operator>(const bign&num)const
{
return num < *this;
}
bool bign::operator<=(const bign&num)const
{
return !(*this>num);
}
bool bign::operator>=(const bign&num)const
{
return !(*this<num);
}
bool bign::operator!=(const bign&num)const
{
return *this > num || *this < num;
}
bool bign::operator==(const bign&num)const
{
return !(num != *this);
}
bign bign::operator+(const bign &num) const
{
if (sign^num.sign)
{
bign tmp = sign ? num : *this;
tmp.sign = 1;
return sign ? *this - tmp : num - tmp;
}
bign result;
result.len = 0;
int temp = 0;
for (int i = 0; temp || i < (max(len, num.len)); i++)
{
int t = s[i] + num.s[i] + temp;
result.s[result.len++] = t % 10;
temp = t / 10;
}
result.sign = sign;
return result;
}
bign bign::operator++()
{
*this = *this + 1;
return *this;
}
bign bign::operator++(int)
{
bign old = *this;
++(*this);
return old;
}
bign bign::operator+=(const bign &num)
{
*this = *this + num;
return *this;
}
bign bign::operator-(const bign &num) const
{
bign b=num,a=*this;
if (!num.sign && !sign)
{
b.sign=1;
a.sign=1;
return b-a;
}
if (!b.sign)
{
b.sign=1;
return a+b;
}
if (!a.sign)
{
a.sign=1;
b=bign(0)-(a+b);
return b;
}
if (a<b)
{
bign c=(b-a);
c.sign=false;
return c;
}
bign result;
result.len = 0;
for (int i = 0, g = 0; i < a.len; i++)
{
int x = a.s[i] - g;
if (i < b.len) x -= b.s[i];
if (x >= 0) g = 0;
else
{
g = 1;
x += 10;
}
result.s[result.len++] = x;
}
result.clean();
return result;
}
bign bign::operator * (const bign &num)const
{
bign result;
result.len = len + num.len;
for (int i = 0; i < len; i++)
for (int j = 0; j < num.len; j++)
result.s[i + j] += s[i] * num.s[j];
for (int i = 0; i < result.len; i++)
{
result.s[i + 1] += result.s[i] / 10;
result.s[i] %= 10;
}
result.clean();
result.sign = !(sign^num.sign);
return result;
}
bign bign::operator*(const int num)const
{
bign x = num;
bign z = *this;
return x*z;
}
bign bign::operator*=(const bign&num)
{
*this = *this * num;
return *this;
}
bign bign::operator /(const bign&num)const
{
bign ans;
ans.len = len - num.len + 1;
if (ans.len < 0)
{
ans.len = 1;
return ans;
}
bign divisor = *this, divid = num;
divisor.sign = divid.sign = 1;
int k = ans.len - 1;
int j = len - 1;
while (k >= 0)
{
while (divisor.s[j] == 0)
j--;
if (k > j) k = j;
char z[MAX_L];
memset(z, 0, sizeof(z));
for(int i = j; i >= k; i--)
z[j - i] = divisor.s[i] + '0';
bign dividend = z;
if (dividend < divid)
{
k--;
continue;
}
int key = 0;
while (divid*key<=dividend) key++;
key--;
ans.s[k] = key;
bign temp = divid*key;
for (int i = 0; i < k; i++)
temp = temp * 10;
divisor = divisor - temp;
k--;
}
ans.clean();
ans.sign = !(sign^num.sign);
return ans;
}
bign bign::operator/=(const bign&num)
{
*this = *this / num;
return *this;
}
bign bign::operator%(const bign& num)const
{
bign a = *this, b = num;
a.sign = b.sign = 1;
bign result, temp = a / b*b;
result = a - temp;
result.sign = sign;
return result;
}
bign bign::pow(const bign& num)const
{
bign result = 1;
for (bign i = 0; i < num; i++)
result = result*(*this);
return result;
}
bign bign::factorial()const
{
bign result = 1;
for(bign i=1;i<=*this;i++)
result *= i;
return result;
}
void bign::clean()
{
if (len == 0) len++;
while (len > 1 && s[len - 1] == '\0')
len--;
}
bign bign::Sqrt()const
{
if(*this<0)return -1;
if(*this<=1)return *this;
bign l=0,r=*this,mid;
while(r-l>1)
{
mid=(l+r)/2;
if(mid*mid>*this)
r=mid;
else
l=mid;
}
return l;
}
bign::~bign()
{
}
bign n,m,t,i,j,k,s(1),res,ans;
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cin>>t;
for(k=1;k<=t;k++)
{
cin>>n>>m;
for(i=1;i<=n;i++)
{
s=s*i;
}
while(s!=0)
{
res=s%10;
if(res==m)
{
ans++;
}
s=s/10;
}
cout<<ans<<endl;
s=1;
ans=0;
}
return 0;
}
普通做法：AC
#include <stdio.h>
#include <iostream>
#include <string.h>
int a[20001],n,m,t,s;//n！，m是数码，t是有t位，s存答案，a数组是一串高精度数
using namespace std;
inline int multe(int n)
{
register int i,j;
for(i=1;i<=t;i++)//先让a数组的每一位都能乘以n
{
a[i]*=n;
}
for(i=1;i<=t;i++)//从1到位数进行循环
{
if(a[i]>9)//如果需要进位
{
a[i+1]+=a[i]/10;//进位
a[i]%=10;
if(i+1>t)//如果现在是最高位
{
t++;//位+1
}
}
}
return 0;
}
inline void lxydl()//lxy大佬太强啦！
{
for(register int i=t;i>=1;i--)//这个倒或正可能无所谓吧，但是高精度后出来的数字都是倒序的，那我们也倒叙
{
if(a[i]==m)//如果当前的a[i]等于数
{
s++;//算符+1
}
}
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
register int i,j,l;//l是l组数据
cin>>l;
while(l--)
{
cin>>n>>m;
s=0;//答案初始化
memset(a,0,sizeof(a));//清除缓存
a[1]=1;//假设第一位数为1，否则0乘任何数都是0
t=1;//至少有1位
for(i=1;i<=n;i++)//求n！
{
multe(i);//a数组乘以i
}
lxydl();//找n!里有多少m这个数
cout<<s<<endl;//输出答案
}
return 0;
}
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