Let the Balloon Rise
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 50554 Accepted Submission(s): 18117
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.
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.
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
1 //定义一个color[100],每次输入一个color之后 2 //与之前的相互比较 如果相同 num++ 3 //最后输出最大的num 4 #include<iostream> 5 #include<string.h> 6 #include<stdio.h> 7 #include<ctype.h> 8 #include<algorithm> 9 #include<stack> 10 #include<queue> 11 #include<set> 12 #include<math.h> 13 #include<vector> 14 #include<map> 15 #include<deque> 16 #include<list> 17 using namespace std; 18 int main() 19 { 20 int n,i,j,num[1000]; 21 int max=0,t=0; 22 char color[1000][16]; 23 while(scanf("%d",&n)!=EOF) 24 { 25 if(n) 26 { 27 num[0]=0; 28 scanf("%s",color[0]); 29 for(i=1;i<n;i++) 30 { 31 num[i]=0; 32 scanf("%s",color[i]); 33 for(j=0;j<i-1;j++) 34 if(strcmp(color[i],color[j])==0) num[i]+=1; 35 } 36 max=0; 37 t=0; 38 for(i=1;i<n;i++) 39 if(max<num[i]) 40 { 41 max=num[i]; 42 t=i;} 43 printf("%s ",color[t]); 44 } 45 } 46 return 0; 47 }