We stack glasses in a pyramid, where the first row has 1 glass, the second row has 2 glasses, and so on until the 100th row. Each glass holds one cup (250ml) of champagne.
Then, some champagne is poured in the first glass at the top. When the top most glass is full, any excess liquid poured will fall equally to the glass immediately to the left and right of it. When those glasses become full, any excess champagne will fall equally to the left and right of those glasses, and so on. (A glass at the bottom row has it’s excess champagne fall on the floor.)
For example, after one cup of champagne is poured, the top most glass is full. After two cups of champagne are poured, the two glasses on the second row are half full. After three cups of champagne are poured, those two cups become full – there are 3 full glasses total now. After four cups of champagne are poured, the third row has the middle glass half full, and the two outside glasses are a quarter full, as pictured below.
Now after pouring some non-negative integer cups of champagne, return how full the j-th glass in the i-th row is (both i and j are 0 indexed.)
Explanation:We poured1cup of champagne tothe top glass of the tower(which isindexed as(0,0)).There will be no excess liquid so all the glasses under the top glass will remain empty.
Explanation:We poured2cups of champagne tothe top glass of the tower(which isindexed as(0,0)).There isone cup of excess liquid.The glass indexed as(1,0)andthe glass indexed as(1,1)will share the excess liquid equally,andeachwill get half cup of champange.
poured will be in the range of [0, 10 ^ 9].
query_glass and query_row will be in the range of [0, 99].
Idea: DP + simulation
define dp[i][j] as the volume of champagne will be poured into the j -th glass in the i-th row, dp[i][j] can be greater than 1.
Init dp = poured.
Transition: if dp[i][j] > 1, it will overflow, the overflow part will be evenly distributed to dp[i+1][j], dp[i+1][j+1]
A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can become B after some number of shifts on A.