[原]CodeForces 478C Table Decorations 思维题

楚东方 16/12/31 16:30:29


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 rg 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 rg 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.

Example
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.





判断最大的是否是其他的两倍

#include <iostream> 
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <fstream>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef  long long ll;
typedef  unsigned long long  ull; 
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};

int main()
{
	ll re = 0;
	ll a[4];
	cin >> a[0]>>a[1]>>a[2];
	sort(a,a+3);
	if(a[0]+a[1]<=a[2]/2)
	{
		re = a[0]+a[1];
	}
	else	
	{
		re = (a[0]+a[1]+a[2])/3;
	}
	cout << re <<endl;
	return 0;
} 



作者:chudongfang2015 发表于2016/12/31 16:30:29 原文链接
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