家人们我知道了高精度乘法和加法，也有他俩的函数，但是不会写阶乘，又没大佬支支招

elong123 @ 2023-05-10 17:31:14

by elong123 @ 2023-05-10 17:32:25

@elong123 看看我的函数

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
char* multiply(char *num1,char * num2){
    int a[2002]={0},b[2002]={0},c[2002]={0};
    int la = strlen(num1);
    int lb = strlen(num2);
    int lc = la + lb;
    //1.倒序存入
    for(int i = 0;i < la;i++){
        a[la - i - 1] = num1[i] - '0';
    }
    for(int j = 0; j <lb ;j++){
        b[lb - j - 1] = num2[j] -'0';
    }
//  //2.乘法的进制存储
//  for(int i = 0; i<la;i++){
//      for(int j = 0 ; j < lb ; j++){
//          c[i + j] += a[i] * b[j];
//          c[i + j + 1] += c[i + j] /10 ;
//          c[i + j] %= 10;
//      }
//  }
    //2.1加法的进制转化
    for(int i = 0; i< lc; i++){
        c[i] += a[i] + b[i];
        c[i + 1] = c[i] /10;
        c[i] %= 10;
    }
    //3.排除特例
    while(c[lc] == 0 && lc>0) lc--;
    //4.定义一个动态字符串存储
    char *re = (char *)malloc(sizeof(char)*(lc+2));
    //5.存储
    for(int i = lc ;i >=0 ; i--){
        re[lc - i] = c[i] +'0';//将数组转为字符串+‘0’；
    }
    re[lc + 1] = 0;
    return re;
}
int main(){
    int n;
    char num1[2002],num2[2002];
    scanf("%d",&n);
    fr
//  char num1[20002],num2[20002];
//  scanf("%s \n %s",num1,num2);
//  char* num = multiply(num1,num2);
//  printf("%s",num);
//  free(num);
//  return 0;

}

by wwb11 @ 2023-05-16 20:19:27

有个阴间的

#include<iostream>
using namespace std;
int main()
{
    string q[60]={"0","1","3","9","33","153","873","5913","46233","409113","4037913","43954713","522956313","6749977113","93928268313","1401602636313","22324392524313","378011820620313","6780385526348313","128425485935180313","2561327494111820313","53652269665821260313","1177652997443428940313","27029669736328405580313","647478071469567844940313","16158688114800553828940313","419450149241406189412940313","11308319599659758350180940313","316196664211373618851684940313","9157958657951075573395300940313","274410818470142134209703780940313","8497249472648064951935266660940313","271628086406341595119153278820940313","8954945705218228090637347680100940313","304187744744822368938255957323620940313","10637335711130967298604907294846820940313","382630662501032184766604355445682020940313","14146383753727377231082583937026584420940313","537169001220328488991089808037100875620940313","20935051082417771847631371547939998232420940313","836850334330315506193242641144055892504420940313","34289376947494122614363304694584807557656420940313","1439295494700374021157505910939096377494040420940313","61854558558074209658512637979453093884758552420940313","2720126133346522977702138448994068984204397080420940313","122342346998826717539665299944651784048588130840420940313","5624964506810915667389970728744906677010239883800420940313","264248206017979096310354325882356886646207872272920420940313","12678163798554051767172643373255731925167694226950680420940313","620960027832821612639424806694551108812720525606160920420940313","31035053229546199656252032972759319953190362094566672920420940313"};
    int n;
    cin>>n;
    cout<<q[n];
    return 0;
}

不推荐使用（但有用）

by wwb11 @ 2023-05-16 20:20:08

网上搜的表

by Sqj147 @ 2023-06-10 17:29:15

// 1 * 2 * 3 * ... * n
a[0] = b[0] = 1;
for (int i = 2; i <= n; i ++) {
    for (j = 0; j < 100; j ++) 
        b[j] *= i;// 1 * 2 * ... * i
    // 高精度处理
    for (j = 0; j < 100; j ++) 
        a[j] += b[j];// 求和
    // 高精度处理
}
// 去除前导0
for (i = 100; i >= 0 && a[i] == 0; i --);
// 倒序输出
for (j = i; j >= 0; j --) cout << a[j];