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  • CF987A Infinity Gauntlet 模拟

    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 n

    (0n6

    ) — the number of Gems in Infinity Gauntlet.

    In next n

    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 m

    (0m6

    ) — the number of absent Gems.

    Then in m

    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
    Input
    Copy
    4
    red
    purple
    yellow
    orange
    Output
    Copy
    2
    Space
    Time
    Input
    Copy
    0
    Output
    Copy
    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<cstdio>
    #include<algorithm>
    #include<cstdlib>
    #include<cstring>
    #include<string>
    #include<cmath>
    #include<map>
    #include<set>
    #include<vector>
    #include<queue>
    #include<bitset>
    #include<ctime>
    #include<deque>
    #include<stack>
    #include<functional>
    #include<sstream>
    //#include<cctype>
    //#pragma GCC optimize(2)
    using namespace std;
    #define maxn 900005
    #define inf 0x7fffffff
    //#define INF 1e18
    #define rdint(x) scanf("%d",&x)
    #define rdllt(x) scanf("%lld",&x)
    #define rdult(x) scanf("%lu",&x)
    #define rdlf(x) scanf("%lf",&x)
    #define rdstr(x) scanf("%s",x)
    typedef long long  ll;
    typedef unsigned long long ull;
    typedef unsigned int U;
    #define ms(x) memset((x),0,sizeof(x))
    const long long int mod = 1e9 + 7;
    #define Mod 1000000000
    #define sq(x) (x)*(x)
    #define eps 1e-3
    typedef pair<int, int> pii;
    #define pi acos(-1.0)
    //const int N = 1005;
    #define REP(i,n) for(int i=0;i<(n);i++)
    typedef pair<int, int> pii;
    inline ll rd() {
        ll x = 0;
        char c = getchar();
        bool f = false;
        while (!isdigit(c)) {
            if (c == '-') f = true;
            c = getchar();
        }
        while (isdigit(c)) {
            x = (x << 1) + (x << 3) + (c ^ 48);
            c = getchar();
        }
        return f ? -x : x;
    }
    
    ll gcd(ll a, ll b) {
        return b == 0 ? a : gcd(b, a%b);
    }
    int sqr(int x) { return x * x; }
    
    
    /*ll ans;
    ll exgcd(ll a, ll b, ll &x, ll &y) {
        if (!b) {
            x = 1; y = 0; return a;
        }
        ans = exgcd(b, a%b, x, y);
        ll t = x; x = y; y = t - a / b * y;
        return ans;
    }
    */
    int n;
    map<string, string>mp;
    map<string, int>mp2;
    string ss[10];
    int main() {
        //ios::sync_with_stdio(0);
        mp["purple"] = "Power"; mp["green"] = "Time";
        mp["blue"] = "Space"; mp["orange"] = "Soul";
        mp["red"] = "Reality"; mp["yellow"] = "Mind";
        rdint(n);
        ss[1] = "red"; ss[2] = "blue"; ss[3] = "yellow";
        ss[4] = "purple"; ss[5] = "green"; ss[6] = "orange";
        for (int i = 1; i <= n; i++) {
            string tmp; cin >> tmp;
            mp2[tmp] = 1;
        }
        cout << 6 - n << endl;
        for (int i = 1; i <= 6; i++) {
            if (!mp2[ss[i]]) {
                cout << mp[ss[i]] << endl;
            }
        }
        return 0;
    }
    
    EPFL - Fighting
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  • 原文地址:https://www.cnblogs.com/zxyqzy/p/10273565.html
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