hduOJ1004 Let the Balloon Rise //算法

Let the Balloon Rise

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 147124    Accepted Submission(s): 58434

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

 

我的想法是用子符串数组记录每种颜色，遍历每个元素与它之后元素相等的次数存储在num[]数组中，比较num数组中最大的值，记录其位置，也就是在字符串数组中出现最多次的位置。

 1 #include<iostream>
 2 #include<cstring>
 3 #include<string>
 4 using namespace std;
 5 int main() {
 6     while(1) {
 7         int n;
 8         cin>>n;
 9         //string *color=new string[n];
10         string color[1002];
11         if(n==0||n>1002)
12             break;
13         else {
14             for(int i=0; i<n; i++)
15                 cin>>color[i];
16         }
17         //int *num=new int[n];
18         int num[1002];
19         memset(num,0,n);
20         for(int i=0; i<n; i++) {            
21             for(int m=i+1; m<n; m++) {
22                 if(color[i]==color[m])
23                     num[i]++;                //记录当前值与之后的值相等个数 
24                     
25             }
26         }
27 
28         int k=0,b=num[0];
29         
30         for(int i=0; i+1<n; i++) {
31             if(b<num[i]) {
32                 b=num[i];
33                 k=i;                    //记录个数最大的位置 
34             }
35         }
36         cout<<color[k]<<endl;
37         memset(num,0,n);
38 //        delete[] color;
39 //        delete[] num;
40     }
41     return 0;
42 }

 

 

 

转载于:https://www.cnblogs.com/yangxin6017/p/9509265.html