• A. Initial Bet(Codeforces Round #273)


    A. Initial Bet
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    There are five people playing a game called "Generosity". Each person gives some non-zero number of coins b as an initial bet. After all players make their bets of b coins, the following operation is repeated for several times: a coin is passed from one player to some other player.

    Your task is to write a program that can, given the number of coins each player has at the end of the game, determine the size b of the initial bet or find out that such outcome of the game cannot be obtained for any positive number of coins b in the initial bet.

    Input

    The input consists of a single line containing five integers c1, c2, c3, c4 and c5 — the number of coins that the first, second, third, fourth and fifth players respectively have at the end of the game (0 ≤ c1, c2, c3, c4, c5 ≤ 100).

    Output

    Print the only line containing a single positive integer b — the number of coins in the initial bet of each player. If there is no such value of b, then print the only value "-1" (quotes for clarity).

    Sample test(s)
    input
    2 5 4 0 4
    
    output
    3
    
    input
    4 5 9 2 1
    
    output
    -1
    
    Note

    In the first sample the following sequence of operations is possible:

    1. One coin is passed from the fourth player to the second player;
    2. One coin is passed from the fourth player to the fifth player;
    3. One coin is passed from the first player to the third player;

    1. One coin is passed from the fourth player to the second player.      

    判一下是否为5的倍数,注意特判全为0

    代码:
    #include <iostream>
    #include <cstdio>
    using namespace std;
    
    int main()
    {
        int a[10];
        int ans=0;
        for(int i=0;i<5;i++)
        {
        scanf("%d",&a[i]);
        ans+=a[i];
        }
        if(ans==0)
        {
            printf("-1
    ");
            return 0;
        }
        if(ans%5==0)
        printf("%d
    ",ans/5);
        else
        printf("-1
    ");
        return 0;
    }
    


    版权声明:本文博客原创文章。博客,未经同意,不得转载。

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  • 原文地址:https://www.cnblogs.com/yxwkf/p/4610250.html
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