CodeForces 478D-Red-Green Towers__codeforces

題目

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 - 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.

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

Note

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

題義

給出r個紅方塊，g個綠方塊，要擺出圖上的那種形狀，就是第i層要有i個，每層都一個顏色，而且要擺出能達到的最大高度,要求能擺出的種數。 解法

設h為最大高度，可以知道h最大是1000不到，設dp[i][j]表示從上往下，擺到第i層，紅方塊還剩j個的方案數，則

dp[i+1][j]=dp[i+1][j]+dp[i][j]//第i+1層不放紅方塊的情況

dp[i+1][j-i-1]=dp[i+1][j-i-1]+dp[i][j]//第i+1層方紅方塊的情況

需要滾動數組最佳化空間 代碼

#include<cstdio>#include<cstring>#define mod 1000000007int dp[2][200005],sum[1005];int main(){int r,g,i,h,j;scanf("%d%d",&r,&g);if(r==0||g==0){printf("1\n");return 0;}sum[0]=0;for(i=1;;i++){sum[i]=sum[i-1]+i;if(sum[i]>r+g)break;}h=i-1;dp[1][r]=1;dp[1][r-1]=1;for(i=1;i<h;i++){int tmp1=i%2,tmp2=(i+1)%2;memset(dp[tmp2],0,sizeof(dp[tmp2]));for(j=0;j<=r;j++){if(dp[tmp1][j]==0)continue;if(g-sum[i]+r-j>=i+1)dp[tmp2][j]=(dp[tmp2][j]+dp[tmp1][j])%mod;if(j>=i+1)dp[tmp2][j-i-1]=(dp[tmp2][j-i-1]+dp[tmp1][j])%mod;}}int ans=0;for(i=0;i<=r;i++)ans=(ans+dp[h%2][i])%mod;printf("%d\n",ans);return 0;}