codeforces 478D

/****************************************************
   file name: cf478D.cpp
   author: Kalis
   creat time: Tue 21 Oct 2014 04:20:40 PM CST
***************************************************/


/********************************************************************************************************** 
  
  定义dp(i,j)为前i层总共用了j个红色方块的方法数，则dp(i,j)可以由两个状态得到：当前层为红，当前层不为红
  状态转移方程为：dp(i,j) = dp(i - 1,j) + dp(i - 1,j - i);

***********************************************************************************************************/   


#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
#include <vector>
#include <queue>

using namespace std;

const int MAXN = 2 * 1e5 + 10;
const int mod = 1e9 + 7;

long long dp[MAXN];
int r,g;

int main()
{
	scanf("%d%d",&r, &g);
	int temp = 0;
	int k = 0;
	while(temp <= r + g)
	{
		k ++;
		temp += k;
	}
	int h = k - 1;
	long long re = ((1 + h) * h >> 1) - g;
	memset(dp,0,sizeof(dp));
	dp[0] = 1;
	long long ans = 0;
	for(int i  = 1; i <= h; i ++)
	{
		int j;
		for(j = r; j >= 0; j --)
		{
			if(j >= i)
				dp[j] = (dp[j] + dp[j - i]) % mod;
			if( i == h && j >= re)
				ans = (dp[j] + ans) % mod;
		}
		
	}
	printf("%I64d\n",ans);
    return 0;
}