HDOJ-1004 Let the Balloon Rise

http://acm.hdu.edu.cn/showproblem.php?pid=1004

输入N个字符串  输出出现频率最高的字符串

# include <stdio.h>
# include <string.h>
# define MAX 1005

struct BALLOON
{
	char Color[20];
	int Times;//同颜色气球出现次数
}Balloon[MAX];
int n, Count;

void Set_Balloon(char Color[])
{
	for(int i = 1; i <= Count; i++)
	{
		//如果这种颜色已经出现过 次数+1
		if(!strcmp(Balloon[i].Color, Color))
		{
			Balloon[i].Times++;
			return ;
		}
	}
	//否则加入表中 次数==1
	strcpy(Balloon[++Count].Color, Color);
	Balloon[Count].Times = 1;
}

int main()
{
	while(scanf("%d",&n) && n)
	{
		char Temp[20];
		Count = 0;
		//逐个读取颜色并放入表中
		for(int i = 1; i <= n; i++)
		{
			scanf("%s",Temp);
			Set_Balloon(Temp);
		}
		//遍历表 找到出现频率最高的颜色
		int Id;
		for(int i = 1, max = -1; i <= Count; i++)
		{
			if(Balloon[i].Times > max)
			{
				max = Balloon[i].Times;
				Id = i;
			}
		}

		printf("%s\n",Balloon[Id].Color);
	}

	return 0;
}

  

时间: 2024-08-09 02:01:12

HDOJ-1004 Let the Balloon Rise的相关文章

hdoj 1004 Let the Balloon Rise(模拟 || 字典树)

Let the Balloon Rise http://acm.hdu.edu.cn/showproblem.php?pid=1004 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 84401    Accepted Submission(s): 31831 Problem Description Contest time again

HDOJ 1004 - Let the Balloon Rise(让气球飞起来)

Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 138223    Accepted Submission(s): 54591 Problem Description Contest time again! How excited it is to see balloons floating ar

HDOJ 1004 Let the Balloon Rise (字符串+stl)

题目: 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 co

HDU 1004 Let the Balloon Rise【STL&lt;map&gt;】

Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 123800    Accepted Submission(s): 48826 Problem Description Contest time again! How excited it is to see balloons floating ar

hdu 1004 Let the Balloon Rise

Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 80469    Accepted Submission(s): 30293 Problem Description Contest time again! How excited it is to see balloons floating aro

杭电1004 Let the Balloon Rise

Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 120507    Accepted Submission(s): 47270 Problem Description Contest time again! How excited it is to see balloons floating ar

HDU 1004 Let the Balloon Rise map

Let the Balloon Rise Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 104108    Accepted Submission(s): 40046 Problem Description Contest time again! How excited it is to see balloons floating ar

HDU 1004 Let the Balloon Rise (map使用)

分析:题意简单,就用map的知识即可 #include <stdio.h> #include <iostream> #include <string.h> #include <map> using namespace std; int main() { int i,a,b,k; int N; char num[10]; while (~scanf("%d",&N)) { map<string,int> mp; whil

水题/hdu 1004 Let the Balloon Rise

题意 给出n个字符串,输出出现次数最多的那个 分析 存下字符串后排序,再统计,输出 Accepted Code 1 /* 2 PROBLEM:hdu1004 3 AUTHER:Nicole Lam 4 MEMO:水题 5 */ 6 7 #include<iostream> 8 #include<cstring> 9 #include<string> 10 #include<algorithm> 11 using namespace std; 12 13 in

HDU ACM 1004 Let the Balloon Rise

用一个数组保存出现的字符串即可,数量用一个整形数组保存. #include<iostream> using namespace std; int main() { char a[1001][20],b[20]; int c[1001]; int n,i,k,j,sum,sumpos; bool find; while(cin>>n && n) { getchar(); memset(c,0,sizeof(c)); k=0; for(i=0;i<n;i++) {