HDUOJ 1004 Let the Balloon Rise（字符串统计水题）

HDUOJ 1004 Let the Balloon Rise（字符串统计水题）

Let the Balloon Rise

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

Total Submission(s): 101430 Accepted Submission(s): 38880

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

5greenredblueredred3pinkorangepink0

Sample Output

redpink

Author

WU, Jiazhi

Source

​​ZJCPC2004​​

Recommend

JGShining   |   We have carefully selected several similar problems for you:  ​​1008​​​ ​​1005​​​ ​​1009​​​ ​​1019​​​ ​​1021​​

AC代码：

#include#include#includeusing namespace std;int main(){ string a[1000]; int T=0,k=0,i,j; while(cin>>T){ if(T==0)break; int b[1000]={0}; for( i=1;i<=T;i++){ // scanf("%s",&a[i]); cin>>a[i]; } for(i=1;i<=T;i++)for( j=1;j<=T;j++){ if(a[i]==a[j]){ b[i]++; } } int maxx=0; for( i=1;i<=T;i++){ if (b[i] > b[maxx]) maxx = i; } // printf("%s",a[max]); cout<

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。