who is the best?
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 128 Accepted Submission(s): 87
Problem Description
There are N people want to choose the best person. Each person select the best person ai,
.John wants to know that who received the most number of votes.
Input
The first line contains a single integer T(1≤T≤50),indicating
the number of test cases.
Each test case begins with an integer N(1≤N≤100),indicating the number of person.
Next N lines contains an integer ai(1≤ai≤N).
Each test case begins with an integer N(1≤N≤100),indicating the number of person.
Next N lines contains an integer ai(1≤ai≤N).
Output
For each case, output an integer means who is the best person. If there are multiple answers, print the minimum index.
Sample Input
2 10 1 2 3 4 5 6 7 8 9 10 5 3 3 3 3 3
Sample Output
1 3
官方题解:
我们对于每个ai都进行计数,即b[ai]++,如此之后,我们可以一个循环语句i=1−>n来寻找最大的bi,注意此时应是bi>MAX,而不是bi≤MAX,这样才能保证出现的是编号最小的。另外需要注意的是每次做完后数组应当清0,否则会影响下次的答案,由于各种非确定性因素我在小数据就已经把没清0的程序卡死了。
代码:
#include <iostream> #include <cstdio> #include <cstring> using namespace std; int a[110]; int main() { int t,n; scanf("%d",&t); while(t--) { scanf("%d",&n); memset(a,0,sizeof(a)); int max=0; for(int i=0;i<n;i++) { int x; scanf("%d",&x); a[x]++; if(a[x]>max) max=a[x]; } int ans; for(int i=0;i<=n;i++) if(a[i]==max) { ans=i; break; } printf("%d\n",ans); } return 0; }
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