A. Infinity Gauntlet

You took a peek on Thanos wearing Infinity Gauntlet. In the Gauntlet there is a place for six Infinity Gems:

the Power Gem of purple color,
the Time Gem of green color,
the Space Gem of blue color,
the Soul Gem of orange color,
the Reality Gem of red color,
the Mind Gem of yellow color.
Using colors of Gems you saw in the Gauntlet determine the names of absent Gems.

Input
In the first line of input there is one integer nn (0≤n≤60≤n≤6) — the number of Gems in Infinity Gauntlet.

In next nn lines there are colors of Gems you saw. Words used for colors are: purple, green, blue, orange, red, yellow. It is guaranteed that all the colors are distinct. All colors are given in lowercase English letters.

Output
In the first line output one integer mm (0≤m≤60≤m≤6) — the number of absent Gems.

Then in mm lines print the names of absent Gems, each on its own line. Words used for names are: Power, Time, Space, Soul, Reality, Mind. Names can be printed in any order. Keep the first letter uppercase, others lowercase.

Examples
inputCopy
4
red
purple
yellow
orange
outputCopy
2
Space
Time
inputCopy
0
outputCopy
6
Time
Mind
Soul
Power
Reality
Space
Note
In the first sample Thanos already has Reality, Power, Mind and Soul Gems, so he needs two more: Time and Space.

In the second sample Thanos doesn’t have any Gems, so he needs all six.

#include <iostream>
#include <string>
#include <map>
using namespace std;

string s[10];
bool vis[7];

void check(string a)
{
    if(a == "purple")
        vis[1] = 1;
    if(a == "green")
        vis[2] = 1;
    if(a == "blue")
        vis[3] = 1;
    if(a == "orange")
        vis[4] = 1;
    if(a == "red")
        vis[5] = 1;
    if(a == "yellow")
        vis[6] = 1;
    return ;
}

int main()
{
    map<int, string> mp;
    mp.insert(pair<int, string> (1, "Power"));
    mp.insert(pair<int, string> (2, "Time"));
    mp.insert(pair<int, string> (3, "Space"));
    mp.insert(pair<int, string> (4, "Soul"));
    mp.insert(pair<int, string> (5, "Reality"));
    mp.insert(pair<int, string> (6, "Mind"));
    int n;
    while(cin >> n)
    {
        for(int i = 0; i < n; ++i)
        {
            cin >> s[i];
            check(s[i]);
        }
        cout << 6 - n << '\n';
        for(int i = 1; i <= 6; ++i)
            if(!vis[i])
            {
                map<int, string>::iterator a = mp.find(i);
                cout << a -> second << '\n';
            }
    }
    return 0;
}

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10-14 23:01
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中国地质大学(武汉) Java
CUG芝士圈:虽然是网上的项目,但最好还是包装一下,然后现在大部分公司都在忙校招,十月底、十一月初会好找一些。最后,boss才沟通100家,别焦虑,我去年暑假找第一段实习的时候沟通了500➕才有面试,校友加油
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