Spoj one0如何避免字符串

spoj onezero how to avoide strings?

本文关键字:字符串 何避免 one0 Spoj      更新时间:2023-10-16

在spoj问题onezero

http://www.spoj.com/problems/ONEZERO/

我在试着做朋友。对于每个状态,我在每个级别上使用余数。在阅读了评论和许多论坛之后,我明白了字符串不能在这里使用。那么如何修改我的代码呢:

#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<cstdio>
#include<ctime>
#include<cctype>
#include<cassert>
#include<climits>
#include<cerrno>
#include<cfloat>
#include<ciso646>
#include<clocale>
#include<csetjmp>
#include<csignal>
#include<cstdarg>
#include<cstddef>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<cwchar>
#include<cwctype>
//containers
#include<vector>
#include<list>
#include<map>
#include<queue>
#include<deque>
#include<set>
#include<complex>
#include<string>
#include<stack>
#include<bitset>
#include<istream>
#include<valarray>
//IOs
#include<iostream>
#include<sstream>
#include<iomanip>
#include<fstream>
#include<exception>
#include<ios>
#include<iosfwd>
#include<ostream>
#include<iterator>
#include<stdexcept>
#include<streambuf>

//algorithm & miscellaneous
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<limits>
#include<locale>
#include<memory>
#include<new>
// My Terms
#define pb push_back
#define mp make_pair
#define ins insert
#define fir first
#define sec second
#define PRINT(x)        cout << #x << " " << x << endl
#define pi acos(-1)
#define ll long long
#define EM empty()
#define sz(a) int((a).size())
#define all(c) (c).begin(),(c).end()
#define fill(a,v)     memset(a, v, sizeof(a))

//http://lab.polygonal.de/?p=81
using namespace std;
int exist[20010];
int main()
{
    int test;
    cin>>test;
    while(test--)
    {
        memset(exist,0,sizeof(int)*20010 );
        int  t;
        cin>>t;
        if(t==1)
        {
            cout<<"1n";
            break;
        }

        int rem=0;
        string num;
        queue < pair<string,int> > q;  // one for number and one for remainder
        q.push(mp("1",1));

        while(q.size())
        {
            num=q.front().first;
            rem=q.front().second;
            exist[rem]=1;
            q.pop();

            if( rem == 0)
            {
                cout<<num<<"n";
                break;
            }

            if(exist[((rem*10 +1)%t)]==0 &&   ((rem*10 +1)%t)<20000 )
            {
                q.push( mp(num+'1',((rem*10 +1)%t)   ) ) ;
                exist[((rem*10 +1)%t)]=1;
            }
            if(exist[((rem*10 +0)%t)]==0 && ((rem*10 +0)%t)<20000  )
            {
                q.push( mp(num+'0',  (rem*10 +0)%t ) );
                exist[ ((rem*10 +0)%t ) ]=1;
            }
        }


    }


}

我如何在这里避免字符串。请告诉我怎么做。

代替push num + '0'和num + '1';你可以只传递一个整数,它是字符串的十进制等效物。例如:代替q.push(mp('1001', (rem*10)%t)),你可以使用q.push(mp(9, (rem*10)%t))。这样你可以获得更多的速度。

但这仍然会给TLE。为了克服这个问题,你应该有一个"访问过的"数组来记住你过去遇到的剩余值,所以你不应该再把它们推到队列中。

通过使用'visited'数组,我得到AC