Let the Balloon RiseTime
limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 85666 Accepted Submission (s): 32330
Problem Descriptioncontest Time again! How excited it was to see balloons floating around. But to tell you a secret, the judges ' favorite time was guessing the most popular problem. When the contest was over, they would count the balloons of each color and find the result.
This is the year that they decide to leave the lovely job to you.
Inputinput contains multiple test cases. Each test case is starts with a number n (0 < N <=) – 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 lower-case letters.
A test Case with N = 0 terminates the-input and this-test case are not-to-be processed.
Outputfor each case, print the color of balloon for the most popular problem to a single line. It is guaranteed this there is a unique solution for each test case.
Sample Input
5greenredblueredred3pinkorangepink0
Sample Output
Redpink
AUTHORWU, Jiazhi
SourceZJCPC2004
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Explain the topic: the main meaning of the topic is multiple case input: Each case the first line is n, contains n data, that is, the color of n strings read into, calculate the most occurrences of the color and output!
The topic is very simple! I've built a string array here in Java string[] Strarray = new String[n]; To save each color, Use an integer a array to hold the number of occurrences of each color. (The subscript of the array of arrays corresponds to the subscript of the string array Strarray) finally calculates the maximum number of occurrences, the output is OK! This is straight straight to the question No turning! Easy accpted Everyone oier 51 Happy Holidays!
Import Java.io.*;import java.util.*;p ublic class main{public static void Main (string[] args) {//TODO auto-generated Metho D stubscanner input = new Scanner (system.in), while (Input.hasnext ()) {int n = input.nextint (); if (n = = 0) Break;input.nextli NE (); string[] Strarray = new String[n];int a[] = new Int[n];for (int i = 0; i < n; i++) {Strarray[i] = Input.nextline ();} for (int i = 0, i < n; i++) {for (int j = 0; J < N; j + +) {if (Strarray[i].compareto (strarray[j]) = = 0) {a[i]++;}}} int flag = 0;int max = 0;for (int i = 0; i < n; i++) {if (A[i] > max) {max = A[i];flag = i;}} System.out.println (Strarray[flag]);}}
Hdu-1004-let the Balloon Rise (directly new a string array compareto!)
