题解 | #N的阶乘#

#include <iostream>
using namespace std;
const int MAX=3000;//试出来的
struct bigInterger {
    int data[MAX];
    int length = 1;
    bigInterger();
    bigInterger(int l);
    bigInterger operator*(bigInterger b);
    bigInterger operator+(bigInterger b);
    void calculateLength();
};
bigInterger::bigInterger() {
    for (int i = 0; i < MAX; i++) {
        data[i] = 0;
    }
}
bigInterger::bigInterger(int l) {
    for (int i = 0; i < MAX; i++) {
        data[i] = 0;
    }
    if (l == 0) return;
    int i = 0;
    while (l != 0) {
        data[i++] = l % 10;
        l /= 10;
    }
    length = i;
}
void bigInterger::calculateLength() {
    int count = 0;
    int i = 99;
    while (data[i--] != 0) {
        count++;
    }
    length = MAX - count;
}
bigInterger bigInterger::operator*(bigInterger b) {
    bigInterger ans = bigInterger();
    ans.length=length + b.length;//乘积的位数最多是乘数位数的和
    for (int i = 0; i < b.length; i++) {
        for (int j = 0; j < length; j++) {
            ans.data[i + j] += b.data[i] * data[j];
        }
    }
    int carry = 0;
    for (int i = 0; i < ans.length; i++) {
        int temp = ans.data[i] + carry;
        ans.data[i] = temp % 10;
        carry = temp / 10;
    }
    for(int i=ans.length-1;i>=0;i--){//计算位数时不算前导零
        if(ans.data[i]==0) ans.length--;
        else break;
    }
    return ans;
}
int main() {
    int n;
    while (cin >> n) { // 注意 while 处理多个 case
        // cout << a + b << endl;
        if (n == 0) {
            cout << 1 << endl;
            continue;
        }
        bigInterger ans = bigInterger(1);
        while (n != 1) {
            ans = ans * bigInterger(n--);
        }
        for (int i = ans.length - 1; i >= 0; i--) {
            cout << ans.data[i];
        }
        cout << endl;
    }
}
// 64 位输出请用 printf("%lld")