Let the Balloon Rise
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 103000 Accepted Submission(s): 39524Problem 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
本题题意：首先输入数据个数n，以下n行输入n个字符串表示气球颜色，输出出现次数最多的颜色。
AC代码：
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 using namespace std;
6
7 const int MAX=1010;
8 int a[MAX];
9 char s[MAX][16];//本想用于最多15种颜色开的二维数组，后来发现一个问题
10 int n;
11
12 int main(){
13 while(scanf("%d",&n)&&n!=0){
14 for(int i=0;i<n;i++){//输入
15 scanf("%s",&s[i]);
16 }
17 memset(a,0,sizeof(a));
18 int ans=0;
19 for(int i=0;i<n-1;i++){
20 for(int j=i+1;j<n;j++){
21 if(strcmp(s[i],s[j])==0){//遇到相等的气球则a[i]++
22 a[i]++;
23 }
24 }
25 if(a[i]>ans)
26 ans=a[i];//ans为最大
27 }
28 for(int i=0;i<n;i++){
29 if(a[i]==ans)
30 puts(s[i]);
31 }
32 }
33 return 0;
34 }
PS：
代码注释中提到的问题，s数组中的[16]看起来并没有用上，可去掉后就会出现编译问题，求大神指教[16]在这里到底做了什么。