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Let the Balloon Rise HDU 1004
Let the Balloon Rise
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 150 Accepted Submission(s) : 63
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.<br><br>This year, they decide to leave this lovely job to you. <br>
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
Author
WU, Jiazhi
#include<iostream> #include<cstdio> #include<cstring> using namespace std; int main(){ char str[1001][16]; int count[1001]; int n,i,j,index; int cnt; while(scanf("%d",&n),n){ index=0; char temp[16]; memset(count,0,sizeof(count)); while(n--){ scanf("%s",temp); for(j=0;j<index;j++){ if(strcmp(temp,str[j])==0){ count[j]++; break; } } if(j==index){ memcpy(str[index++],temp,sizeof(temp)); count[j]++; } } for(i=0,cnt=0;i<index;i++) if(count[i]>count[cnt]) cnt=i; printf("%s\n",str[cnt]); } }
Let the Balloon Rise HDU 1004
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