#include <iostream>
int main(){
    int q[6],a[6]={1,1,2,2,2,8};
    for(int i=0;i<6;i++)std::cin>>q[i];
    for(int i=0;i<6;i++)std::cout<<a[i]-q[i]<<" ";
    return 0;}

使用数组存储每种棋子的正确数量，然后依次输入并做差即可。

#include <bits/stdc++.h>
using namespace std;
int a[7] = {0, 1, 1, 2, 2, 2, 8}, x;
int main()
{
    for (int i = 1; i <= 6; i++) {
        cin >> x;
        cout << a[i] - x << " " ;
    }
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
string s=" 112228";
int x;
int main(){
	for (int i=1;i<=6;i++){scanf("%d",&x);printf("%d ",s[i]-'0'-x);}
}

# 菜题
#include <bits/stdc++.h>
using namespace std;
int main()
{
    int a[6];
    int b[6];
    int c[6];
    a[0]=1;
    a[1]=1;
    a[2]=2;
    a[3]=2;
    a[4]=2;
    a[5]=8;
    for (int i=0; i<6; i++)
    {
        cin >>b[i];
    }
    for (int i=0; i<6; i++)
    {
        if (b[i]<a[i])
        {
            c[i]=a[i]-b[i];
        }
        else
        {
            c[i]=a[i]-b[i];
        }
    }
    for (int i=0; i<6; i++)
    {
        cout <<c[i]<<" ";
    }
    return 0;
}

#include <bits/stdc++.h>//by hetao4185102
using namespace std;
int a, b, c, d, e, f;
int main()
{
cin >> a >> b >> c >> d >> e >> f;
if(a != 1) cout << 1 - a << " ";
else cout << 0 << " ";
if(b != 1) cout << 1 - b << " ";
else cout << 0 << " ";
if(c != 2) cout << 2 - c << " ";
else cout << 0 << " ";
if(d != 2) cout << 2 - d << " ";
else cout << 0 << " ";
if(e != 2) cout << 2 - e << " ";
else cout << 0 << " ";
if(f != 8) cout << 8 - f << " ";
else cout << 0 << " ";
return 0;
}

#include <bits/stdc++.h>//by hetao4185102
using namespace std;
int a, b, c, d, e, f;
int main()
{
cin >> a >> b >> c >> d >> e >> f;
if(a != 1) cout << 1 - a << " ";
else cout << 0 << " ";
if(b != 1) cout << 1 - b << " ";
else cout << 0 << " ";
if(c != 2) cout << 2 - c << " ";
else cout << 0 << " ";
if(d != 2) cout << 2 - d << " ";
else cout << 0 << " ";
if(e != 2) cout << 2 - e << " ";
else cout << 0 << " ";
if(f != 8) cout << 8 - f << " ";
else cout << 0 << " ";
return 0;
}

这代码只有前8行是有用的，水印爱要不要。欢迎白嫖(皮

#include <bits/stdc^^.h>
using namespace std;
void ImW();
int n,chess[7]={0,1,1,2,2,2,8},input[7];//这个你都看不懂你还是别学了
signed main(void){
    for(int i=1;i<=6;i++)    cin>>input[i];
    for(int i=1;i<=6;i++)    cout<<chess[i]-input[i]<<' ';
}
void ImW()
{
    /*我是水印*/
}

#include <bits/stdc++.h>
using namespace std;
int x, b[10]{ 0, 1, 1, 2, 2, 2, 8 };
int main()
{
    for (int i = 1; i <= 6; i++)
    {
        cin >> x;
        cout << b[i] - x << " ";
    }
    return 0;
}

这题很简单，暴力计算就好

#include <bits/stdc++.h>//by hetao4185102
using namespace std;
int a, b, c, d, e, f;
int main()
{
cin >> a >> b >> c >> d >> e >> f;
if(a != 1) cout << 1 - a << " ";
else cout << 0 << " ";
if(b != 1) cout << 1 - b << " ";
else cout << 0 << " ";
if(c != 2) cout << 2 - c << " ";
else cout << 0 << " ";
if(d != 2) cout << 2 - d << " ";
else cout << 0 << " ";
if(e != 2) cout << 2 - e << " ";
else cout << 0 << " ";
if(f != 8) cout << 8 - f << " ";
else cout << 0 << " ";
return 0;
}

这个问题的解决思路是先计算出每种棋子需要添加或删除的数量，然后将结果输出。

对于国王和皇后，由于每套正确的棋子只能有一个，所以如果输入的数量大于1，则需要删除多余的棋子，即数量减去1。如果数量为0，则需要添加一个棋子，即1减去数量。

对于车、象和马，由于每套正确的棋子都需要两个，所以如果输入的数量大于2，则需要删除多余的棋子，即数量减去2。如果数量小于2，则需要添加缺少的棋子，即2减去数量。

对于兵，由于每套正确的棋子需要8个，所以如果输入的数量大于8，则需要删除多余的棋子，即数量减去8。如果数量小于8，则需要添加缺少的棋子，即8减去数量。

以上就是求解这个问题的思路，根据输入的棋子数量和上述规则，计算出每种棋子需要添加或删除的数量，并将结果输出。

#include <iostream>

int main() {
    int king, queen, rook, bishop, knight, pawn;
    std::cin >> king >> queen >> rook >> bishop >> knight >> pawn;

    int missingKing = 1 - king;
    int missingQueen = 1 - queen;
    int missingRook = 2 - rook;
    int missingBishop = 2 - bishop;
    int missingKnight = 2 - knight;
    int missingPawn = 8 - pawn;

    std::cout << missingKing << " " << missingQueen << " " << missingRook << " "
              << missingBishop << " " << missingKnight << " " << missingPawn << std::endl;

    return 0;
}