Codeforces Round #273 (Div. 2)

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.

搞了好久才发现题目读错了,题目要求是每桌三个气球的颜色不能所有一样

给出r，b。g分别代表红色，蓝色，绿色气球的数量，每三个气球能够布置一张桌子。要求同一个桌子上的三个气球颜色不能全一样，求最多能布置多少张桌子

先把r,g,b从小到达排序,如果a<b<c,如果c>=2*(a+b),那么我们不断从c中取出2个,a,b中随便取出一个,则a+b能够所实用完,没有最优的了,这样的情况就是a+b种,如果c < 2*(a+b),如果如今仅仅有a,b两种气球,那么我们假有x种rgg,y种rrg,那么有(2x+y) + (2y+x) = a+b,解得x+y = (a+b)/3,那么我们大胆如果如果是三种的话则答案是(a+b+c)/3,事实上这是成立的,能够把最少的那种加到中间的那种中变成两种

#include <iostream>
#include <cstdio>
#include <cstdio>
#include <algorithm>
using namespace std;

int main()
{
    long long a[3];
    scanf("%I64d%I64d%I64d",&a[0],&a[1],&a[2]);
    sort(a,a+3);
    long long ans=0;
    if((a[0]+a[1])<=a[2]/2)
    {
       ans=a[0]+a[1];
    }
    else
    {
        ans=(a[0]+a[1]+a[2])/3;
    }
    printf("%I64d
",ans);
    return 0;
}

There are r red and g green blocks for construction of the red-green tower. Red-green tower can be built following next rules:

Red-green tower is consisting of some number of levels;
Let the red-green tower consist of n levels, then the first level of this tower should consist of n blocks, second level — of n - 1 blocks, the third one — of n - 2 blocks, and so on — the last level of such tower should consist of the one block. In other words, each successive level should contain one block less than the previous one;
Each level of the red-green tower should contain blocks of the same color.

$\text{[math]}$

Let h be the maximum possible number of levels of red-green tower, that can be built out of r red and g green blocks meeting the rules above. The task is to determine how many different red-green towers having h levels can be built out of the available blocks.

Two red-green towers are considered different if there exists some level, that consists of red blocks in the one tower and consists of green blocks in the other tower.

You are to write a program that will find the number of different red-green towers of height h modulo 109 + 7.

Input

The only line of input contains two integers r and g, separated by a single space — the number of available red and green blocks respectively (0 ≤ r, g ≤ 2·105, r + g ≥ 1).

Output

Output the only integer — the number of different possible red-green towers of height h modulo 109 + 7.

Sample test(s)

input
4 6

output
2

input
9 7

output
6

input
1 1

output
2

Note

The image in the problem statement shows all possible red-green towers for the first sample.

建造一个有r个红木块，g个绿木块的“红绿塔”。

红绿塔的最下层有k个木块，上一层有k-1个木块……直到最上层有一个木块，每层的木块颜色要同样，求这些木块能堆出的最高的“红绿塔”的种数。（r+g个木块不一定全都要用上）

设dp[i][j]为前i层使用了j个红色砖块时的方案数，则dp[i][j]=dp[i-1][j]（即新的一层所有为绿色）+dp[i-1][j-i]（即新的一层所有为红色），答案为所以dp[h]的和

i最多到900。j最多到2·10⁵，显然数组不能开到这么大。所以须要使用滚动数组(这个地方类似01背包的将二维数组缩小为一维数组)

上面二维能够转换成dp[j] = dp[j] + dp[j-i],假设不懂的能够先去看看01背包那里是怎样将二维转一维的

include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<string>
#define MOD 1000000007
int dp[200005];
using namespace std;
typedef long long LL;

int main()
{

    int r,g;
    while(cin>>r>>g)
    {
        memset(dp,0,sizeof(dp));
        int h;
        for( h = 1; h*(h+1)/2 <= r+g; h++);//先求最多有多少层
        h--;
        dp[0] = 1;//这代表没有红色方块的1-h层的方案永远仅仅有1种。即所有都是绿色的必然是1种了
        for(int i = 1; i <= h ; i++)
        {
            for(int j = r; j >= 0; j --)
            {
                if(j - i>= 0)
                {
                    dp[j]  = (dp[j]  + dp[j -i])%MOD;
                }
            }
        }
        LL Max = h*(h+1)/2 , ans = 0;
        for(int i = 0; i <= r; i++)
        {
            if(Max - i <= g)     ans = (ans + dp[i])%MOD;//Max-i的意思是里面除去红色的方块总共同拥有多少绿色方块,假设小于g,则该方案不陈立
        }
        cout<<ans<<endl;

    }
    return 0;
}