题解 | #N的阶乘#

#include <iostream>
#include <string>
using namespace std;
string multiply(string a, int b) {
    int size = a.size();
    int carry = 0;
    for (int i = size - 1; i >= 0 ;
            i--) {//从低位到高位逐位相乘，取模，进位
        int temp = (a[i] - '0') * b + carry;
        a[i] = temp % 10 + '0';
        carry = temp / 10;

    }
    //最终的进位可能很大
    while (carry != 0) {
        a = char(carry % 10 + '0') + a;
        carry /= 10;
    }
    return a;
}
int main() {
    int N;
    while (cin >> N) { // 注意 while 处理多个 case
        // cout << a + b << endl;
        string ans = to_string(N);
        for (int i = N - 1; i >= 2; i--) {
            ans = multiply(ans, i);
        }
        cout << ans << endl;
    }
}
// 64 位输出请用 printf("%lld")