codeforces Round #273(div2) C解题报告

C. Table Decorations

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

You have r red, g green and b blue
balloons. To decorate a single table for the banquet you need exactly three balloons. Three balloons attached to some table shouldn't have the same color. What maximum number t of
tables can be decorated if we know number of balloons of each color?

Your task is to write a program that for given values r, g and b will
find the maximum number t of tables, that can be decorated in the required manner.

Input

The single line contains three integers r, g and b (0 ≤ r, g, b ≤ 2·109)
— the number of red, green and blue baloons respectively. The numbers are separated by exactly one space.

Output

Print a single integer t — the maximum number of tables that can be decorated in the required manner.

Sample test(s)

input

5 4 3

output

4

input

1 1 1

output

1

input

2 3 3

output

2

Note

In the first sample you can decorate the tables with the following balloon sets: "rgg", "gbb",
"brr", "rrg", where "r",
"g" and "b" represent the red, green and blue balls, respectively.

题目大意：

给出三种不同颜色的气球‘r', 'g', 'b'，装饰在桌子旁，要求每个桌子的三个气球颜色不能一样。

解法：

三个气球的个数按照从大到小排序，然后如果最大的不能以abb的方式匹配其他两个，就可以用总数直接除3。

PS：排列组合题，这一块比较弱，说不出所以然出来，自己之前也是各种贪心各种WA，看了题解才知道，原来是这样=_=#。

代码：

#include <iostream>
#include <algorithm>
#define LL long long

using namespace std;

LL a[10];
LL ans;

void init() {
cin >> a[1] >> a[2] >> a[3];
sort(a+1, a+4);
}

void solve() {
if (a[3] >= 2*(a[1]+a[2]))
ans = a[1]+a[2];
else
ans = (a[1]+a[2]+a[3])/3;

cout << ans << endl;
}

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