D - Red-Green Towers dp

D - Red-Green Towers

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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 - 1blocks, 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.

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 Input

Input

4 6

Output

2

Input

9 7

Output

6

Input

1 1

Output

2

Hint

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

dp,我的代码是枚举红色球的层数，当第i层为红色是，i层上面有[0,r]个 红色的，可推出dp[i+j]=dp[i+j]+dp[j],最后

再统计红色的个数就行了，红色最少为max(h*(h+1)/2-g,0)。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
const int mod=1e9+7;
long long dp[500000];
int main()
{
  int r,g;
  cin>>r>>g;
  int t=r+g;
  int h=sqrt(t*2);
  while(h*(h+1)/2>t)
  h--;
  dp[0]=1;
  for(int i=1;i<=h;i++)
    for(int j=r;j>=0;j--)
    {
     dp[i+j]=(dp[i+j]+dp[j])%mod;
    }
    long long ans=0;
    for(int i=max(h*(h+1)/2-g,0);i<=r;i++)
    {
      ans=(ans+dp[i])%mod;


    } printf("%lld\n",ans);
    return 0;
}

题解：这一题是动归。用D[i][j]表示从上往下第i层，红色方块用掉j个时的方案数。那么D[i][j]=D[i-1][j]+D[i-1][j-i]，数组可以滚动压缩压成D[j]，而时间上看似过不了，其实是可以过的。（分析确实是2*10^8，但是这是算满最大的情况，实际情况中有很多都没算，所以跑出来只有1S）

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1. #include <iostream>  
2. #include <cstring>  
3. #include <cstdio>  
4. #include <cmath>  
5.   
6. using namespace std;  
7.   
8. long long d[1000000],ans;  
9. int h,n,m;  
10.   
11. int main()  
12. {  
13.     scanf("%d%d",&n,&m);  
14.     h=(int) sqrt(2*(n+m));  
15.     while (h*(h+1)>2*(n+m)) h--;  
16.     if (n) d[1]=1;  
17.     if (m) d[0]=1;  
18.     for (int i=2;i<=h;i++)  
19.         for (int j=(i+1)*i/2;j>=0;j--)  
20.     {  
21.         if (i*(i+1)/2-j>m) d[j]=0;  
22.         if (j<=n && j-i>=0) d[j]=(d[j-i]+d[j])%1000000007;  
23.     }  
24.     ans=0;  
25.     for (int i=0;i<=n;i++) ans=(ans+d[i])%1000000007;  
26.     printf("%I64d",ans);  
27.     return 0;  
28. }