codeforces Round #273(div2) A解题报告

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:

One coin is passed from the fourth player to the second player;

One coin is passed from the fourth player to the fifth player;

One coin is passed from the first player to the third player;

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

题目大意：

每个人在一开始都拥有一样的金币数量b，给出赌完钱之后的每个人剩下的金币，求出b。

解法：

直接加起来看是否可以整除

代码：

#include <cstdio>

int sum, c[10];

void init() {
for (int i = 1; i <= 5; i++) {
scanf("%d", &c[i]);
sum += c[i];
}
}

void solve() {
if (sum % 5 != 0 || sum == 0)
printf("-1\n");
else
printf("%d\n", sum/5);
}

int main() {
init();
solve();
}