基础练习 | 阶乘

Posted 2020-10-19 TQCAI

花了一下午鼓捣了高精度加法函数。

一下代码中multi函数用string做容器，存在溢出的问题，可改为vector<int>或vector<long long>

#include <stdio.h>
#include <memory.h>
#include <math.h>
#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <algorithm>
#include <map>


#define I scanf
#define OL puts
#define O printf
#define F(a,b,c) for(a=b;a<c;a++)
#define FF(a,b) for(a=0;a<b;a++)
#define FG(a,b) for(a=b-1;a>=0;a--)
#define LEN 10000
#define MAX 0x06FFFFFF
#define V vector<int>

using namespace std;

string multi(string a,string b);

int main() {
//    freopen("d:/input/A1130/1.txt","r",stdin);
    int n,i;
    scanf("%d",&n);
    string ans="1";
    F(i,2,n+1){
        char cnum[1000];
        sprintf(cnum,"%d",i);
        string num(cnum);
        ans=multi(ans,num);
    }
    printf("%s\n",ans.c_str());
//    printf("%s\n",multi("50","60").c_str());

    return 0;
}

string multi(string a,string b){
    int an=a.size(),bn=b.size(),n=an+bn,cnt=0;
    int i,j;
    FF(i,an) a[i]-=48;
    FF(i,bn) b[i]-=48;
    string ans(n,0);
    string* add=new string[bn];
    for ( i=0 ; i<bn ; i++ ){
        int t=b[bn-1-i];
        add[i]=string(n,0);
        string &s=add[i];
        for ( j=0 ; j<an ; j++ ){
            s[j+i]+=a[an-1-j]*t;
            if(s[j+i]>9){
                int d=s[j+i]/10;
                s[j+i]-=d*10;
                s[j+i+1]+=d;
            }
        }
    }
    for ( i=0 ; i<n ; i++ ){
        for ( j=0 ; j<bn ; j++ ){
            ans[i]+=add[j][i];
        }
        if(ans[i]>9){
            int d=ans[i]/10;
            ans[i]-=d*10;
            ans[i+1]+=d;
        }
        ans[i]+=48;
    }
//    FF(i,n) O("%d ",ans[i]-48);
//    OL("");
    for ( i=n-1 ; i>=0 ; i-- ){
        if(ans[i]!=48) break;
    }
    ans=ans.substr(0,i+1);
    reverse(ans.begin(),ans.end());
    return ans;
}