100! 对于64 位整数来说太大了。它有158 数字。您必须实现 BigInteger 库。希望@LightOj Judge 创建者@Jane Alom Jan 有一个很好的实现，你可以检查一下。我正在分享他的实现，您可以针对此问题进行修改和测试。
#include <cstdio>
#include <string>
#include <algorithm>
#include <iostream>
using namespace std;
struct Bigint {
// representations and structures
string a; // to store the digits
int sign; // sign = -1 for negative numbers, sign = 1 otherwise
// constructors
Bigint() {} // default constructor
Bigint( string b ) { (*this) = b; } // constructor for string
// some helpful methods
int size() { // returns number of digits
return a.size();
}
Bigint inverseSign() { // changes the sign
sign *= -1;
return (*this);
}
Bigint normalize( int newSign ) { // removes leading 0, fixes sign
for( int i = a.size() - 1; i > 0 && a[i] == '0'; i-- )
a.erase(a.begin() + i);
sign = ( a.size() == 1 && a[0] == '0' ) ? 1 : newSign;
return (*this);
}
// assignment operator
void operator = ( string b ) { // assigns a string to Bigint
a = b[0] == '-' ? b.substr(1) : b;
reverse( a.begin(), a.end() );
this->normalize( b[0] == '-' ? -1 : 1 );
}
// conditional operators
bool operator < ( const Bigint &b ) const { // less than operator
if( sign != b.sign ) return sign < b.sign;
if( a.size() != b.a.size() )
return sign == 1 ? a.size() < b.a.size() : a.size() > b.a.size();
for( int i = a.size() - 1; i >= 0; i-- ) if( a[i] != b.a[i] )
return sign == 1 ? a[i] < b.a[i] : a[i] > b.a[i];
return false;
}
bool operator == ( const Bigint &b ) const { // operator for equality
return a == b.a && sign == b.sign;
}
// mathematical operators
Bigint operator + ( Bigint b ) { // addition operator overloading
if( sign != b.sign ) return (*this) - b.inverseSign();
Bigint c;
for(int i = 0, carry = 0; i<a.size() || i<b.size() || carry; i++ ) {
carry+=(i<a.size() ? a[i]-48 : 0)+(i<b.a.size() ? b.a[i]-48 : 0);
c.a += (carry % 10 + 48);
carry /= 10;
}
return c.normalize(sign);
}
Bigint operator - ( Bigint b ) { // subtraction operator overloading
if( sign != b.sign ) return (*this) + b.inverseSign();
int s = sign; sign = b.sign = 1;
if( (*this) < b ) return ((b - (*this)).inverseSign()).normalize(-s);
Bigint c;
for( int i = 0, borrow = 0; i < a.size(); i++ ) {
borrow = a[i] - borrow - (i < b.size() ? b.a[i] : 48);
c.a += borrow >= 0 ? borrow + 48 : borrow + 58;
borrow = borrow >= 0 ? 0 : 1;
}
return c.normalize(s);
}
Bigint operator * ( Bigint b ) { // multiplication operator overloading
Bigint c("0");
for( int i = 0, k = a[i] - 48; i < a.size(); i++, k = a[i] - 48 ) {
while(k--) c = c + b; // ith digit is k, so, we add k times
b.a.insert(b.a.begin(), '0'); // multiplied by 10
}
return c.normalize(sign * b.sign);
}
Bigint operator / ( Bigint b ) { // division operator overloading
if( b.size() == 1 && b.a[0] == '0' ) b.a[0] /= ( b.a[0] - 48 );
Bigint c("0"), d;
for( int j = 0; j < a.size(); j++ ) d.a += "0";
int dSign = sign * b.sign; b.sign = 1;
for( int i = a.size() - 1; i >= 0; i-- ) {
c.a.insert( c.a.begin(), '0');
c = c + a.substr( i, 1 );
while( !( c < b ) ) c = c - b, d.a[i]++;
}
return d.normalize(dSign);
}
Bigint operator % ( Bigint b ) { // modulo operator overloading
if( b.size() == 1 && b.a[0] == '0' ) b.a[0] /= ( b.a[0] - 48 );
Bigint c("0");
b.sign = 1;
for( int i = a.size() - 1; i >= 0; i-- ) {
c.a.insert( c.a.begin(), '0');
c = c + a.substr( i, 1 );
while( !( c < b ) ) c = c - b;
}
return c.normalize(sign);
}
// output method
void print() {
if( sign == -1 ) putchar('-');
for( int i = a.size() - 1; i >= 0; i-- ) putchar(a[i]);
}
};
int main() {
Bigint a, b, c; // declared some Bigint variables
/////////////////////////
// taking Bigint input //
/////////////////////////
string input; // string to take input
cin >> input; // take the Big integer as string
a = input; // assign the string to Bigint a
cin >> input; // take the Big integer as string
b = input; // assign the string to Bigint b
//////////////////////////////////
// Using mathematical operators //
//////////////////////////////////
c = a + b; // adding a and b
c.print(); // printing the Bigint
puts(""); // newline
c = a - b; // subtracting b from a
c.print(); // printing the Bigint
puts(""); // newline
c = a * b; // multiplying a and b
c.print(); // printing the Bigint
puts(""); // newline
c = a / b; // dividing a by b
c.print(); // printing the Bigint
puts(""); // newline
c = a % b; // a modulo b
c.print(); // printing the Bigint
puts(""); // newline
/////////////////////////////////
// Using conditional operators //
/////////////////////////////////
if( a == b ) puts("equal"); // checking equality
else puts("not equal");
if( a < b ) puts("a is smaller than b"); // checking less than operator
return 0;
}
由于问题的源限制为 2000 字节，因此添加洞 BigInteger 库将超过源限制。
所以这里只能进行字符串乘法。
下面给出了一个干净的乘法方法，使用C++
#include<bits/stdc++.h>
using namespace std;
int main(){
int t;
cin >> t;
while(t--){
vector<int> arr;
int n;
cin >> n;
arr.push_back(1);
int carry = 0;
for(int i = 2; i <= n; i++){
vector<int> t;
for(int j = arr.size() - 1; j >= 0; j--){
int r = arr[j] * i + carry;
carry = r / 10;
t.push_back(r % 10);
}
while(carry){
t.push_back(carry % 10);
carry /= 10;
}
reverse(t.begin(), t.end());
arr = t;
}
for(auto el : arr){
cout << el;
}
cout << endl;
}
return 0;
}
输入
2
10
100
输出
3628800
93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000
还添加了一个简单的python 实现
test = int(input())
for i in range(0, test):
n = int(input())
res = 1
for j in range(2, n + 1):
res = res * j
print(res)