BestCoder Round #50 (div.2) 1001
2015-08-08 21:19
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Problem Description
AFA want to distribution her money to somebody.She divide her money into n same parts.One who want to get the money can get more than one part.But if one man's money is more than the sum of all others'.He shoule be punished.Each one who get a part of money
would write down his ID on that part.
Input
There are multiply cases.For each case,there is a single integer n(1<=n<=1000) in first line.In second line,there are n integer a1,a2...an(0<=ai<10000)ai is the the ith man's ID.
Output
Output ID of the man who should be punished.If nobody should be punished,output -1.
Sample Input
Sample Output
AFA want to distribution her money to somebody.She divide her money into n same parts.One who want to get the money can get more than one part.But if one man's money is more than the sum of all others'.He shoule be punished.Each one who get a part of money
would write down his ID on that part.
Input
There are multiply cases.For each case,there is a single integer n(1<=n<=1000) in first line.In second line,there are n integer a1,a2...an(0<=ai<10000)ai is the the ith man's ID.
Output
Output ID of the man who should be punished.If nobody should be punished,output -1.
Sample Input
3 1 1 2 4 2 1 4 3
Sample Output
1 -1
#include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <stack> #include <algorithm> #include <cmath> #define LL long long #define INF 0x3f3f3f3f using namespace std; int a[10010]; int ha[10110]; int main() { int n; while(cin>>n) { memset(ha,0,sizeof(ha)); for(int i=0;i<n;i++) { cin>>a[i]; ha[a[i]]++; } int pos = -1; for(int i=0;i<n;i++) { if(ha[a[i]] > n/2) { pos = a[i]; break; } } cout<<pos<<endl; } return 0; }
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