zoukankan      html  css  js  c++  java
  • BC #50 1001 Distribution money

    Distribution money

     
     Accepts: 713
     
     Submissions: 1881
     Time Limit: 2000/1000 MS (Java/Others)
     
     Memory Limit: 65536/65536 K (Java/Others)
    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
    3
    1 1 2
    4
    2 1 4 3
    
    Sample Output
    1
    -1
     1 #include <stdio.h>
     2 #include <string.h>
     3 
     4 int main()
     5 {
     6     int a[10005];
     7     int n,i,j,k,flg;
     8     while(scanf("%d",&n)!=EOF)
     9     {
    10         flg=0;
    11         memset(a,0,sizeof(a));
    12         for(i=1;i<=n;i++)
    13         {
    14             scanf("%d",&k);
    15             a[k]++;
    16         }
    17         for(i=0;i<10000;i++)
    18         {
    19             if(a[i]>(n-a[i]))
    20             {
    21                 flg=1;
    22                 j=i;
    23                 break;
    24             }
    25         }
    26         if(flg==1)
    27             printf("%d
    ",j);
    28         else
    29             printf("-1
    ");
    30     }
    31     return 0;
    32 }
    View Code
  • 相关阅读:
    mysql 约束条件 外键 forigen key 介绍
    【洛谷P4655】Building Bridges
    【CF1139D】Steps to One
    【YbtOJ#20073】钻石守卫
    【YbtOJ#20072】相似子串
    【YbtOJ#20071】礼物购买
    【洛谷P4149】Race
    【洛谷P2059】卡牌游戏
    【CF140C】New Year Snowmen
    【GMOJ4282】平方数游戏
  • 原文地址:https://www.cnblogs.com/cyd308/p/4771334.html
Copyright © 2011-2022 走看看