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  • codeforces305A

    Strange Addition

     CodeForces - 305A 

    Unfortunately, Vasya can only sum pairs of integers (ab), such that for any decimal place at least one number has digit 0 in this place. For example, Vasya can sum numbers 505 and 50, but he cannot sum 1 and 4.

    Vasya has a set of k distinct non-negative integers d1, d2, ..., dk.

    Vasya wants to choose some integers from this set so that he could sum any two chosen numbers. What maximal number of integers can he choose in the required manner?

    Input

    The first input line contains integer k (1 ≤ k ≤ 100) — the number of integers.

    The second line contains k distinct space-separated integers d1, d2, ..., dk (0 ≤ di ≤ 100).

    Output

    In the first line print a single integer n the maximum number of the chosen integers. In the second line print n distinct non-negative integers — the required integers.

    If there are multiple solutions, print any of them. You can print the numbers in any order.

    Examples

    Input
    4
    100 10 1 0
    Output
    4
    0 1 10 100
    Input
    3
    2 70 3
    Output
    2
    2 70

    sol:我太菜了,只会分类讨论一大堆情况qaq,简直就是个智障
    #include <bits/stdc++.h>
    using namespace std;
    typedef int ll;
    inline ll read()
    {
        ll s=0;
        bool f=0;
        char ch=' ';
        while(!isdigit(ch))
        {
            f|=(ch=='-'); ch=getchar();
        }
        while(isdigit(ch))
        {
            s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
        }
        return (f)?(-s):(s);
    }
    #define R(x) x=read()
    inline void write(ll x)
    {
        if(x<0)
        {
            putchar('-'); x=-x;
        }
        if(x<10)
        {
            putchar(x+'0'); return;
        }
        write(x/10);
        putchar((x%10)+'0');
        return;
    }
    #define W(x) write(x),putchar(' ')
    #define Wl(x) write(x),putchar('
    ')
    const int N=105;
    int n,a[N],Id[10]={0};
    inline void put0(int n)
    {
        int i;
        for(i=1;i<=n;i++) W(0);
    }
    int main()
    {
        int i,j,ans=0;
        R(n);
        for(i=1;i<=n;i++)
        {
            R(a[i]);
            if(a[i]==0) ans++;//000
            else if(a[i]<10) Id[1]=i;//00x
            else if(a[i]<100)
            {
                if(a[i]%10==0) Id[2]=i;//0x0
                else Id[3]=i; //0xx
            }
            else
            {
                if(a[i]%100==0) Id[4]=i;//x00
                else if(a[i]%10==0) Id[5]=i; //xx0;
                else if((a[i]-a[i]%10)%100==0) Id[6]=i; //x0x
                else Id[7]=i; //xxx
            }
        }
        if(Id[1])
        {
            if(Id[2])
            {
                if(Id[4])
                {
                    Wl(ans+3); put0(ans); W(a[Id[1]]); W(a[Id[2]]); W(a[Id[4]]);
                }
                else
                {
                    Wl(ans+2); put0(ans); W(a[Id[1]]); W(a[Id[2]]);
                }
            }
            else
            {
                if(Id[4])
                {
                    Wl(ans+2); put0(ans); W(a[Id[1]]); W(a[Id[4]]);
                }
                else if(Id[5])
                {
                    Wl(ans+2); put0(ans); W(a[Id[1]]); W(a[Id[5]]);
                }
                else
                {
                    Wl(ans+1); put0(ans); W(a[Id[1]]);
                }
            }
        }
        else if(Id[2])
        {
            if(Id[4])
            {
                Wl(ans+2); put0(ans); W(a[Id[2]]); W(a[Id[4]]);
            }
            else if(Id[6])
            {
                Wl(ans+2); put0(ans); W(a[Id[2]]); W(a[Id[6]]);
            }
            else
            {
                Wl(ans+1); put0(ans); W(a[Id[2]]);
            }
        }
        else if(Id[4])
        {
            if(Id[3])
            {
                Wl(ans+2); put0(ans); W(a[Id[4]]); W(a[Id[3]]);
            }
            else
            {
                Wl(ans+1); put0(ans); W(a[Id[4]]);
            }
        }
        else if(Id[3])
        {
            Wl(ans+1); put0(ans); W(a[Id[3]]);
        }
        else if(Id[5])
        {
            Wl(ans+1); put0(ans); W(a[Id[5]]);
        }
        else if(Id[6])
        {
            Wl(ans+1); put0(ans); W(a[Id[6]]);
        }
        else if(Id[7])
        {
            Wl(ans+1); put0(ans); W(a[Id[7]]);
        }
        else
        {
            Wl(ans); put0(ans);
        }
        return 0;
    }
    /*
    Input
    4
    100 10 1 0
    Output
    4
    0 1 10 100 
    
    Input
    3
    2 70 3
    Output
    2
    2 70 
    */
    View Code
     
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  • 原文地址:https://www.cnblogs.com/gaojunonly1/p/10753277.html
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