8/9，集合的运算

就是暴力的四种方法都算出答案，然后进行比较有没有这个答案就好了

#include<bits/stdc++.h>
#define sf scanf
#define scf(x) scanf("%d",&x)
#define scff(x,y) scanf("%d%d",&x,&y)
#define scfff(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define pf printf
#define prf(x) printf("%d\n",x)
#define mm(x,b) memset((x),(b),sizeof(x))
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
#define pb push_back
typedef long long ll;void read(int &x){
int f=1;x=0;char s=getchar();
while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
while(s>='0'&&s<='9'){x=x*10+s-'0';s=getchar();}
x*=f;}
const ll mod=1e9+100;
const double eps=1e-8;
using namespace std;
const double pi=acos(-1.0);
const int inf=0xfffffff;
const int N=1005;
vector<int> v[3];
vector<int> vv[3];
vector<int> ans[4];
int main()
{
    int x,n,m,k;
    scfff(n,m,k);
    rep(i,0,n)
    {
        scf(x);
        v[0].pb(x);
    }
    rep(i,0,m)
    {
        scf(x);
        v[1].pb(x);
    }
    rep(i,0,k)
    {
        scf(x);
        v[2].pb(x);
    }
    sort(v[0].begin(),v[0].end());
    sort(v[1].begin(),v[1].end());
    sort(v[2].begin(),v[2].end());
    vv[0].pb(v[0][0]);
    vv[1].pb(v[1][0]);
    vv[2].pb(v[2][0]);
    rep(j,0,3)
    rep(i,1,v[j].size())
    {
        if(v[j][i]!=v[j][i-1])
            vv[j].pb(v[j][i]);
    }
    n=vv[0].size();
    m=vv[1].size();
    k=vv[2].size();
    int pos1=0,pos2=0;
    while(1)//进行并集和交集的
    {
        if(pos1==n&&pos2==m)break;
        if(pos1==n)
            ans[0].pb(vv[1][pos2++]);
        else if(pos2==m)
            ans[0].pb(vv[0][pos1++]);
        else 
        {
            if(vv[0][pos1]>vv[1][pos2])
            {
                ans[0].pb(vv[1][pos2++]);
            }else if(vv[0][pos1]<vv[1][pos2])
            {
                ans[0].pb(vv[0][pos1++]);
            }else
            {
                ans[0].pb(vv[0][pos1]);
                ans[1].pb(vv[0][pos1]);
                pos1++;
                pos2++;
            }
        }
    }
    pos1=0;pos2=0;
    while(1)//进行1减2
    {
        if(pos1==n)
            break;
        else if(pos2==m)
            ans[2].pb(vv[0][pos1++]);
        else 
        {
            if(vv[0][pos1]>vv[1][pos2])
            {
                pos2++;
            }else if(vv[0][pos1]<vv[1][pos2])
            {
                ans[2].pb(vv[0][pos1++]);
            }else
            {
                pos1++;
                pos2++;
            }
        }
    }
    pos1=0;pos2=0;
    while(1)//进行2减1
    {
        if(pos2==m)
            break;
        else if(pos1==n)
            ans[3].pb(vv[1][pos2++]);
        else 
        {
            if(vv[0][pos1]>vv[1][pos2])
            {
                ans[3].pb(vv[1][pos2++]);
                
            }else if(vv[0][pos1]<vv[1][pos2])
            {
                pos1++;
            }else
            {
                pos1++;
                pos2++;
            }
        }
    }
    rep(i,0,4)//比较四种答案
    {
        if(ans[i].size()==vv[2].size())
        {
            int temp=1;
            rep(j,0,ans[i].size())
            {
                if(ans[i][j]!=vv[2][j])
                {
                    temp=0;
                    break;
                }
            }
            if(temp)
            {
                rep(j,0,vv[2].size())
                {
                    if(j==0)
                    pf("%d",vv[2][j]);
                    else
                    pf(" %d",vv[2][j]);
                }
                return 0;
            }
        }
    }
    pf("What a pity!");
    return 0;
}