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  • A. Transmigration

    A. Transmigration

    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    In Disgaea as in most role-playing games, characters have skills that determine the character's ability to use certain weapons or spells. If the character does not have the necessary skill, he cannot use it. The skill level is represented as an integer that increases when you use this skill. Different character classes are characterized by different skills.

    Unfortunately, the skills that are uncommon for the given character's class are quite difficult to obtain. To avoid this limitation, there is the so-called transmigration.

    Transmigration is reincarnation of the character in a new creature. His soul shifts to a new body and retains part of his experience from the previous life.

    As a result of transmigration the new character gets all the skills of the old character and the skill levels are reduced according to the k coefficient (if the skill level was equal to x, then after transmigration it becomes equal to [kx], where [y] is the integral part of y). If some skill's levels are strictly less than 100, these skills are forgotten (the character does not have them any more). After that the new character also gains the skills that are specific for his class, but are new to him. The levels of those additional skills are set to 0.

    Thus, one can create a character with skills specific for completely different character classes via transmigrations. For example, creating a mage archer or a thief warrior is possible.

    You are suggested to solve the following problem: what skills will the character have after transmigration and what will the levels of those skills be?

    Input

    The first line contains three numbers nm and k — the number of skills the current character has, the number of skills specific for the class into which the character is going to transmigrate and the reducing coefficient respectively; n and m are integers, and k is a real number with exactly two digits after decimal point (1 ≤ n, m ≤ 20, 0.01 ≤ k ≤ 0.99).

    Then follow n lines, each of which describes a character's skill in the form "name exp" — the skill's name and the character's skill level: name is a string and exp is an integer in range from 0 to 9999, inclusive.

    Then follow m lines each of which contains names of skills specific for the class, into which the character transmigrates.

    All names consist of lowercase Latin letters and their lengths can range from 1 to 20characters, inclusive. All character's skills have distinct names. Besides the skills specific for the class into which the player transmigrates also have distinct names.

    Output

    Print on the first line number z — the number of skills the character will have after the transmigration. Then print z lines, on each of which print a skill's name and level, separated by a single space. The skills should be given in the lexicographical order.

    Examples

    input

    5 4 0.75
    axe 350
    impaler 300
    ionize 80
    megafire 120
    magicboost 220
    heal
    megafire
    shield
    magicboost

    output

    6
    axe 262
    heal 0
    impaler 225
    magicboost 165
    megafire 0
    shield 0

    题意:玩过DNF的人都知道,每个角色到了18级就可以转职,换成一个新的职业,有新的技能。而这个题的意思是在未转职之前的技能点数大于或者等于100的,可以保留下来。而转职之后获得新的技能,如果那个技能在之前已经有点数了(这个点数必定是大于或等于100),那就不需要进行操作,如果没有的话,就将该技能点数赋值为0。本题有一个问题,就是会卡你精度,我就是因为这个测试数据19总是过不了,其他的就没什么注意的了。
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<map>
    
    #define eps 1e-6
    
    using namespace std;
    
    map<string,int> mp;
    
    int main()
    {
        int n,m;
        double k;
        cin>>n>>m>>k;
        string name;
        int exp;
        for(int i=0;i<n;i++)
        {
            cin>>name>>exp;
            exp=exp*k+eps;    //注意:这里是卡精度,要加上行1e-6才能过
            if(exp>=100)    //判断是否大于或等于100
                mp[name]=exp;
        }
        for(int i=0;i<m;i++)
        {
            cin>>name;
            if(mp[name]==0)
                mp[name]=0;
        }
        cout<<mp.size()<<endl;
        for(map<string,int>::iterator it=mp.begin();it!=mp.end();it++)        //map容器里是按字典序存放的
            cout<<it->first<<" "<<it->second<<endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/buhuiflydepig/p/10635400.html
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