HDU/HDOJ 1004 Let the Balloon Rise（颜色统计）

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Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 107654    Accepted Submission(s): 41780

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

题意：

很明确的题意，就是统计出出现最多的颜色。

思路：

没学习结构体排序之前还真不知道怎么做，

不知道把颜色首字母转换为数字用一位数组对不对（毕竟存在BUG不知道颜色都是哪些）。

Code：

#include<stdio.h>#include<stdlib.h>#include<string.h>#include<math.h>#include<algorithm>#define MYDD 1103using namespace std;struct Q {char color[16];} balloons[MYDD];bool cmp_color(Q x,Q y) {//对气球颜色进行排序return strcmp(x.color,y.color)>0? 1:0;}int MAX(int x,int y) {return x<y? x:y;//}int main() {int n;char dd[16];//答案颜色while(scanf("%d",&n)&&n) {for(int j=0; j<n; j++)scanf("%s",balloons[j].color);sort(balloons,balloons+n,cmp_color);char now[16];//记录当前待比较的颜色int max=0;//记录最多气球颜色的个数int ans=0;//记录当前气球颜色的个数strcpy(now,balloons[0].color);strcpy(dd,balloons[0].color);for(int j=0; j<n; j++) {if(strcmp(now,balloons[j].color)==0)ans++;else {//颜色不同if(ans>max) {strcpy(dd,balloons[j-1].color);//ans 记录的是前一个气球颜色的个数max=ans;}strcpy(now,balloons[j].color);//极记为当前气球颜色ans=1;}}puts(dd);}return 0;}/*7C A B A B C C3a B C*/