标签：sse multiple start 色值 mission temp sample other secret
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 149425 Accepted Submission(s): 59368
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
第一种实现方式：结构体比较：
#include<iostream> #include<string.h> using namespace std; struct node { char balloon[20]; int count_ball=0; }; int main() { int N; struct node Color[1005],Temp; while (cin >> N) { if (N == 0) break; int i, j; for (i = 0; i < N; i++) { cin >> Color[i].balloon; Color[i].count_ball = 1; } for (i = 0; i < N; i++)//从当前的颜色开始计数，直到后面的颜色全部统计完，则为该颜色的数量 { for (j = i + 1; j < N; j++) { if (strcmp(Color[i].balloon, Color[j].balloon) == 0) { Color[i].count_ball++; } } } for (i = 0; i < N - 1; i++)//比较，求出最大的颜色 { for (j = i; j < N - 1 - i; j++) { if (Color[j].count_ball > Color[j + 1].count_ball) { Temp = Color[j]; Color[j] = Color[j + 1]; Color[j + 1] = Temp; } } } cout <<Color[i].balloon << endl; } return 0; }
第二种实现方式Map：
#include<iostream> #include<string> #include<map> using namespace std; int main() { int N; map<string, int> ballsum; string str; while (cin >> N&&N > 0) { ballsum.clear();//清空map while (N--) { cin >> str; ballsum[str]++; } int max = 0; string maxclor; map<string, int>::iterator iter;//定义迭代器 for (iter = ballsum.begin(); iter != ballsum.end(); iter++)//迭代查找 { if (iter->second > max)//寻找最大的颜色数量 { max = (*iter).second; maxclor = (*iter).first;//将map中颜色值赋值给maxcolor } } cout << maxclor << endl; } return 0; }
标签：sse multiple start 色值 mission temp sample other secret
原文地址：https://www.cnblogs.com/gcter/p/9735840.html
