大数阶乘

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以前有道作业是大数阶乘, 重新捡起来实现一下, 已经完全理解了该怎么做, 实现细节也明白了.

一会用记事本手工写一个试试.

为了做这个题，还总结了一个斯特林公式取对数估算n!输出结果字符串字符数量的经验公式 ^_^

/// @file src_n_factorial.cpp/// @brief 算n!的实现#include <stdlib.h>#include <stdio.h>#include <math.h>#include <conio.h>#include <memory.h>#include <time.h>#ifndef WORD#define WORD unsigned short#endifvoid calcFact(size_t n); ///< 算n!size_t getFactNCharCnt(size_t n); ///< 估算n!结果字符数int main(int argc, char* argv[]) {    size_t n = 0;    clock_t tmBegin = 0;    clock_t tmEnd = 0;    while (1) {        printf("\nplease input n to fact:");        scanf("%u", &n);                if (0 == n)            break;                printf("\ncalcFact(%u) = ", n);        tmBegin = clock();        calcFact(n);        tmEnd = clock();        printf("cost time = %dS\n", (tmEnd - tmBegin) / CLOCKS_PER_SEC);    }    /// 已经验证过了, 和Windows计算器的n!(n最大3248)比对过    /// 如果UI上显示不下, 可以用 my.exe > d:\txt 来运行程序    /** run result          please input n to fact:100              calcFact(100) = 9332621544394415268169923885626670049071596826438162146859296389        52175999932299156089414639761565182862536979208272237582511852109168640000000000        00000000000000        nResultCharCnt = 160        cost time = 0S              please input n to fact:0                END, press any key to quit    */    printf("\nEND, press any key to quit\n");    _getch();    return 0;}void calcFact(size_t n) {    size_t nArySize = getFactNCharCnt(n); ///< n!结果占用的字符数量    size_t nIndex = 0;    size_t nMul = 0;    size_t nSum = 0;    size_t nUnitMax = 10000; ///< 一个运算单位中能放的最大数.    size_t nResultCharCnt = 0;    WORD wCarry = 0;    WORD* pAry = NULL;    WORD* pTail = NULL;    WORD* pHead = NULL;    WORD* pCur = NULL;    bool bFirstPrint = true;    nArySize = (nArySize - (nArySize % sizeof(WORD))) / sizeof(WORD) + 1;    pAry = (WORD*)malloc(nArySize * sizeof(WORD));    if (NULL == pAry)        return;    memset(pAry, 0, nArySize * sizeof(WORD));    /// 预置了被乘数为1, 新的乘数从2到n    *(pAry + nArySize - 1) = 1;    for (nIndex = 2, pTail = pHead = (pAry + nArySize - 1);        nIndex <= n;        nIndex++) {            /// 用新的乘数从低位字节到高位字节, 一直乘到有进位的最高被乘数字节.            /// 每次的乘积都加上进位            for (pCur = pTail; pCur >= pHead; pCur--) {                nMul = nIndex * (*pCur) + wCarry;                *pCur = nMul % nUnitMax;                wCarry = (nMul - *pCur) / nUnitMax;            }            /// 连续累加进位到高字节            while (wCarry > 0) {                pHead--;                nSum = (*pHead) + wCarry;                (*pHead) = nSum % nUnitMax;                wCarry = (nSum - *(pHead)) / nUnitMax;            }    }    /// 找到第一个不为0的元素位置    pHead = pAry;    while (pHead <= pTail) {        if (0 == *pHead)            pHead++;        else            break;    }    while (pHead <= pTail) {        if (bFirstPrint) {            bFirstPrint = false;            nResultCharCnt = pTail - pHead + 1;            printf("%d", *pHead); ///< 不打印起始数字的开始0字符        }        else             printf("%4.4d", *pHead); ///< 打印0        pHead++;    }    printf("\n");    /// n!结果字符串, 一共打印出来多少字符    printf("nResultCharCnt = %d\n", nResultCharCnt * sizeof(WORD) * 2);    free(pAry);}size_t getFactNCharCnt(size_t n) {    double N = (double)n;    const double E = 2.71828;    return (size_t)(N * (log10(N) - log10(E)) + 4);}

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