HDU1004 Let the ballon raise(水题，字符串的处理)（map简单用法）

LBJHan

于 2017-09-02 09:19:28 发布

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本文链接：https://blog.youkuaiyun.com/Sirius_han/article/details/77793410

本文介绍了一个简单的编程问题——统计比赛中最流行的题目所对应的气球颜色。通过读取一系列颜色输入并使用数组或映射来跟踪每种颜色出现的次数，最终输出出现次数最多的颜色。

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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): 126197 Accepted Submission(s): 49894

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

找出现最多的颜色，每输入一个颜色与前边输入的比较，是否有一样的颜色，有就该颜色++，没有就新建一个颜色；

字符串的比较是strcmp（s1, s2）函数；如果s1=s2返回0，s1<s2返回负值，s1>s2返回正值；

字符串复制是strcpy（s1,s2）函数；把s2复制到s1中；

代码如下：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <string.h>

using namespace std;

int main(){

    char color[1000][20];
    int n;
    int cnt[1000];
    while(scanf("%d",&n), n!=0){
    memset(cnt, 0, sizeof(cnt));

    int i=0;
    char s[20];
    for(int j=0; j<n; j++){
        scanf("%s",s);
        int k;
        for(k=0; k<i; k++){
            if(strcmp(color[k], s)==0) break;
        }
        if(k>=i){
            cnt[i]++;
            strcpy(color[i++], s);//该颜色没有出现过，就新建一个颜色，把s  copy到  color中；
        }
        else cnt[k]++;
    }
    int maxx=0, ans=0;
    for(int j=0; j<i; j++){
        if(maxx<cnt[j]){
            ans=j;
            maxx=cnt[j];
        }
    }
    printf("%s\n",color[ans]);
    }
    return 0;
}

加一个用map做的；

#include <stdio.h>
#include <map>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;


int main(){
	
	int n;
	while(1){
	map<string, int> mp;
	cin >> n;
	if(n==0) break;
	for(int i=1; i<=n; i++){
		string k;
		cin >> k;
		mp[k]++;
	}
	map<string ,int>::iterator come = mp.begin(), go = mp.end(), it;
	int maxx=0;
	for( ; come != go; ++come){
		if(maxx<come->second){
			maxx=come->second;
			it=come;
		}
		//cout << come->first << ' ' << come->second << endl;
	}
	cout << it->first << endl;
	}
	return 0;
}