zoukankan      html  css  js  c++  java
  • hdu 1004 Let the Balloon Rise strcmp、map、trie树

    Let the Balloon Rise

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

    Problem Description
    Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result.

    This year, they decide to leave this lovely job to you. 
     
    Input
    Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) -- the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters.

    A test case with N = 0 terminates the input and this test case is not to be processed.
     
    Output
    For each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.
     
    Sample Input
    5 green red blue red red 3 pink orange pink 0
     
    Sample Output
    red pink
     
    Author
    WU, Jiazhi
    题意:找出出现次数最多的那个字符串;
    strcmp:
    #include<stdio.h>
    #include<string.h>
    char a[1000][20];
    int b[1011];
    int main ()
    {
        int x,y,z,j,k,i,t,max;
        while(scanf("%d",&x)!=EOF)
        {
            if(x==0)
            break;
            getchar();
            memset(b,0,sizeof(b));
            for(i=0;i<x;i++)
            scanf("%s",a[i]);
            for(i=0;i<x;i++)
            for(t=0;t<x;t++)
            if(strcmp(a[i],a[t])==0)
            b[i]++;
            max=b[0];
            k=0;
            for(i=1;i<x;i++)
            if(max<b[i])
            {max=b[i];k=i;}
            printf("%s
    ",a[k]);
        }
        return 0;
    }

    map:

    #include<stdio.h>
    #include<map>
    #include<iostream>
    #include<string.h>
    #include<string>
    using namespace std;
    int main()
    {
        int x,y,z,i,t,max;
        string a,b;
        map<string,int>p;
        while(scanf("%d",&x)!=EOF)
        {
            getchar();
            p.clear();
            if(x==0) break;
            max=0;
            for(i=0;i<x;i++)
            {
                cin>>a;
                p[a]++;
                if(p[a]>max)
                {
                    max=p[a];
                b=a;}
            }
            cout<<b<<endl;
        }
        return 0;
    }

    trie树:

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<string>
    #include<queue>
    #include<algorithm>
    #include<stack>
    #include<cstring>
    #include<vector>
    #include<list>
    #include<set>
    #include<map>
    using namespace std;
    #define ll long long
    #define mod 1000000007
    #define pi (4*atan(1.0))
    const int N=1e5+10,M=5e6+10,inf=1e9+10;
    int a[N][27],sum[M],len,ans=0;
    void init()
    {
        memset(a,0,sizeof(a));
        memset(sum,0,sizeof(sum));
        len=1;
        ans=0;
    }
    char aaa[N];
    int getnum(char a)
    {
        return a-'a';
    }
    void insertt(char *aa)
    {
        int u=0,n=strlen(aa);
        for(int i=0; i<n; i++)
        {
            int num=getnum(aa[i]);
            if(!a[u][num])
            {
                a[u][num]=len++;
            }
            u=a[u][num];
            sum[u]++;
        }
        if(sum[u]>ans)
        {
            ans=sum[u];
            strcpy(aaa,aa);
        }
    }
    int getans(char *aa)
    {
        int u=0,x=strlen(aa);
        for(int i=0; i<x; i++)
        {
            int num=getnum(aa[i]);
            if(!a[u][num])
                return 0;
            u=a[u][num];
        }
        return sum[u];
    }
    char ch[N];
    int main()
    {
        int x,y,z,i,t;
        while(~scanf("%d",&x))
        {
            if(x==0)break;
            init();
            for(i=0; i<x; i++)
            {
                scanf("%s",ch);
                insertt(ch);
            }
            printf("%s
    ",aaa);
        }
        return 0;
    }
     
  • 相关阅读:
    jenkins master-slave配置
    ansible-playbook 变量(vars)
    ansible-playbook && Roles && include
    ansible 循环与条件判断when
    python 微信爬虫实例
    JavaScript: 零基础轻松学闭包
    【干货】用大白话聊聊JavaSE — ArrayList 深入剖析和Java基础知识详解(二)
    【干货】用大白话聊聊JavaSE — ArrayList 深入剖析和Java基础知识详解(一)
    【大结局】《从案例中学习JavaScript》之酷炫音乐播放器(四)
    Java 实现批量重命名,亲测可用(精简版)
  • 原文地址:https://www.cnblogs.com/jhz033/p/5581141.html
Copyright © 2011-2022 走看看