codeforces Round #273(div2) D解题报告

D. Red-Green Towers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 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.

题目大意：

给出绿砖个数和红砖个数，来堆一个塔（如图）。最顶层为1个，按顺序下来就是2, 3 ... n-2, n-1, n。每一层的颜色必须相同，求出最大层数中不同的塔的个数。

解法：

如若用暴力搜索，肯定是不行的，因为状态数太多，h最大是800+，每层有2种状态，暴搜是指数级，这太可怕了。

我们可以确定的就是，给了我们红砖个数和绿砖个数，我们可以求出总数，就可以只用确定其中一个状态，就可以确定另一个状态。

       假设：i为目前在哪一层，j为红砖数量，通过sum-cnt[i-1]-j我们可以得知绿砖还有多少，接下来就是很简单的DP思路了。

给出状态转移方程：

       f[i][j] = f[i-1][j]+f[i-1][j+h[i]];  

代码：

#include <cstdio>#define modn 1000000007int r, g, sum, ans, cnt[1010];int f[2][4*123456];void init() {scanf("%d%d", &r, &g);sum = r+g;f[0][r] = 1;}void solve() {int now=0;int pre=1;for (int i = 1; i <= 1000; i++) {now ^= 1;pre = (now+1)%2; cnt[now] = cnt[pre]+i;if (cnt[now] > sum) {for (int j = 0; j <= r; j++) {ans += f[pre][j];if (ans >= modn)  ans -= modn;}break;}for (int j = 0; j <= r; j++) {f[now][j] = 0;if (sum-cnt[pre]-j >= i)f[now][j] = f[pre][j];f[now][j] += f[pre][j+i];if (f[now][j] >= modn)  f[now][j] -= modn;}}printf("%d\n", ans);}int main() {init();solve();}