数据存储类型。
就是1024！的结果，用什么数据类型来存储呢？ 试过double，long double， 但是结果不行，最后的结果是1，顺便问下，这是为什么。
试过100！结果，后面0的个数高达145个，
那该怎么去验证1024！后面0的个数呢？
我的想法是，把1024！的结果求出，然后每与10求余一次，结果为0 的话，就记录一次。
以上就是问题，和我的解题思路。
请大虾们帮忙看下。
[解决办法]
这个问题，最简单最明了的作法就是gmp大数库或者bc库。
[解决办法]
哥给你整一个吧：
#include <iostream>#include <string>#include <time.h>using namespace std; int COMPARE(string number1, string number2){ int j; int length1 = number1.size(); int length2 = number2.size(); if(number1.size() == 0) number1 = "0"; if(number2.size() == 0) number2 = "0"; j = 0; for(int i = 0; i < length1; ++i) { if(number1[i] == '0') ++j; else break; } number1 = number1.substr(j); j = 0; for(int i = 0; i < length2; ++i) { if(number2[i] == '0') ++j; else break; } number2 = number2.substr(j); length1 = number1.size(); length2 = number2.size(); if(length1 > length2) { return 1; } else if(length1 == length2) { if(number1.compare(number2) > 0) { return 1; } else if(number1.compare(number2) == 0) { return 0; } else { return -1; } } else { return -1; } return 0;}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 MINUS(string number1,string number2){ int i; string result = ""; int length1 = number1.size(); int length2 = number2.size(); if(COMPARE(number2,number1) > 0) { return "-" + MINUS(number2, number1); } reverse(number1.begin(),number1.end()); reverse(number2.begin(),number2.end()); for(i = 0; i < length1 && i < length2; i++) { char c = number1[i] - number2[i] + 48; result = result + c; } if(i < length1) { for(; i < length1; i++) { result = result + number1[i]; } } int carry = 0; for(i = 0; i < (int)result.length(); i++) { int value = result[i] - 48 + carry; if(value < 0) { value = value + 10; carry = -1; } else carry = 0; result[i]=(char)(value + 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;} // 缺省地，商数向下取整，floatpoint用于指定保留小数点的位数string DIVIDE(string number1, string number2, int floatpoint = 0){ int i, j, pos; string result = ""; string tempstr = ""; int length1 = number1.size(); int length2 = number2.size(); if((COMPARE(number2, number1) > 0) && (floatpoint == 0)) { return "0"; } tempstr = number1.substr(0, length2); pos = length2 - 1; while(pos < length1) { int quotient = 0; while(COMPARE(tempstr, number2) >= 0) { quotient++; tempstr = MINUS(tempstr, number2); } result = result + (char)(quotient + 48); pos++; if(pos < length1) { tempstr += number1[pos]; } } if(floatpoint > 0) { result += '.'; string stmp = "1"; int itmp = 0; for(int k = 0; k < floatpoint; ++k) { stmp += '0'; if(COMPARE(MULTIPLY(MINUS(number1, MULTIPLY(DIVIDE(number1, number2), number2)), stmp), number2) < 0) { result += '0'; ++itmp; } } string temp = DIVIDE(MULTIPLY(MINUS(number1, MULTIPLY(DIVIDE(number1, number2), number2)), stmp), number2); if(temp[0] != '0') result += temp; } j = result.size(); for(i = 0; i < j; i++) { if(result[i] != '0') break; } result = result.substr(i, j); return result;}string FACTORIAL(string number){ if(number == "0") return "1"; return MULTIPLY(FACTORIAL(MINUS(number, "1")), number);} int main(int argc, char* argv[]){ clock_t start, end; string number = "1024"; int interval = 0; start = clock(); string result = FACTORIAL(number); end = clock(); interval = (end - start) * 1000 / CLOCKS_PER_SEC; string reverse_result = result; reverse(reverse_result.begin(), reverse_result.end()); int i = 0; while(reverse_result[i] == '0') { ++i; } cout << "Result of 1024!:" << endl; cout << result << endl; cout << "Which contains " << i << " zeroes at the end" << endl; cout << "It took " << interval << "ms" << endl; return 0;}