萌新求带，请问有什么局限性或者是错误吗，为什么一直不能完全通过，求大佬指错

[hsbwyebd]

 

[forever74]

题干的前六个字“用高精度计算”。
这意味着任何内置的数据类型都不足以正确表达题目中需要的数值，您需要用数组保存完整的数据，并自行实现基于数组的加法和阶乘运算函数。
50的阶乘是30414093201713378043612608166064768844377641568960512000000000000 ，
long long放不下的。

[赵4老师]

仅供参考：

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
inline int COMPARE(string str1,string str2) {//相等返回0，大于返回1，小于返回-1
         if (str1.size()>str2.size()) return 1; //长度长的整数大于长度小的整数
    else if (str1.size()<str2.size()) return -1;
    else                              return str1.compare(str2); //若长度相等，则头到尾按位比较
}
string PLUS(string number1,string number2) {
    int i;
    int length1 = number1.size();
    int length2 = number2.size();

    string result="";

    reverse(number1.begin(), number1.end());
    reverse(number2.begin(), number2.end());

    for(i = 0; i < length1 && i < length2; i++) {
        char c = (char)(number1[i] + number2[i] - 48);
        result = result + c;
    }

    while(i < length1) {
        result = result + number1[i];
        ++i;
    }

    while(i < length2) {
        result = result + number2[i];
        ++i;
    }

    int carry = 0;
    for(i = 0; i < (int)result.size(); ++i) {
        int value = result[i] - 48 + carry;
        result[i] = (char)(value % 10 + 48);
        carry = value / 10;
    }

    if(carry !=0 ) {
        result = result + (char)(carry + 48);
    }

    for(i = result.size() - 1; i >= 0; i--) {
        if(result[i] != '0') break;
    }

    result = result.substr(0, i + 1);

    reverse(result.begin(), result.end());
    if(result.length() == 0) result = "0";
    return result;
}
string MULTIPLY(string number1, string number2) {
    int i, j;
    int *iresult;
    int length1 = number1.size();
    int length2 = number2.size();
    string result = "";

    reverse(number1.begin(), number1.end());
    reverse(number2.begin(), number2.end());

    iresult = (int*)malloc(sizeof(int) * (length1 + length2 + 1));
    memset(iresult, 0, sizeof(int) * (length1 + length2 + 1));

    for(i = 0; i < length1; i++) {
        for(j = 0; j < length2; j++) {
            iresult[i+j] += ((number1[i] - 48) * (number2[j] - 48));
        }
    }

    int carry = 0;
    for(i = 0; i < length1 + length2; i++) {
        int value = iresult[i] + carry;
        iresult[i] = value % 10;
        carry = value / 10;
    }

    for(i = length1 + length2 - 1; i >= 0; i--) {
        if(iresult[i] != 0)break;
    }

    for(; i >= 0; i--) {
        result = result + (char)(iresult[i]+48);
    }

    free(iresult);

    if(result == "") result = "0";
    return result;
}
string factorial(string n) {
    string temp = "1";
    string i;
    for(i = "1"; COMPARE(i, n) <= 0; i = PLUS(i, "1")) {
        temp = MULTIPLY(temp, i);
    }
    return temp;
}
int main(void) {
    cout << factorial("100") << endl;
    return 0;
}
//93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000
//

[日立奔腾浪潮微软松下联想]

老赵的千年老代码还是有些价值的~