Codeforces

原创于 2016-07-25 21:06:48 发布 · 367 阅读

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探讨如何使用给定数量的红绿方块构建最大高度的红绿塔，通过动态规划求解不同构建方案的数量。

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暑期集训4 B题

传送门：http://codeforces.com/problemset/problem/478/D

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.

Examples

input
4 6

output
2

input
9 7

output
6

input
1 1

output
2

题意：给你a个红方块和b个绿方块

现让你组成一个三角形，第i层有i个方块，且这一层方块颜色要一样

现问你对于给出的a，b，你所能构造出的最大的三角形中，一共有几种方法

题解：dp题，用dp[i]表示用了i个方块所能构造的红色为最下面一层的三角形的个数

因此初始dp[0]=1

而状态转移则为dp[i+j]=dp[i+j]+dp[j]，i为第i层，j为0到a的任意值

最后答案即为红色能出现的所有情况

其实是网上看到的题解...

看了好久才看懂...

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<iostream>
using namespace std;

int main()
{
	int a,b,i,j,n,ans,sum;
	int dp[400005];
	
	scanf("%d%d",&a,&b);
	sum=a+b;
	ans=0;
	for(i=1;1;i++)
	{
		ans+=i;
		if(ans>sum) break;
	}
	ans=ans-i;
	i--;
	n=i;
	dp[0]=1;
	for(i=1;i<=n;i++)
		for(j=a;j>=0;j--)
			dp[i+j]=(dp[i+j]+dp[j])%1000000007;
	sum=0;
    for(i=max(ans-b,0);i<=a;i++)  
    sum=(sum+dp[i])%1000000007;  
    printf("%d\n",sum);
	return 0;
}