Vases and Flowers Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2635 Accepted Submission(s): 1026
Problem Description
Alice is so popular that she can receive many flowers everyday. She has N vases numbered from 0 to N-1. When she receive some flowers, she will try to put them in the vases, one flower in one vase. She randomly choose the vase A and try to put a flower in the vase. If the there is no flower in the vase, she will put a flower in it, otherwise she skip this vase. And then she will try put in the vase A+1, A+2, ..., N-1, until there is no flower left or she has tried the vase N-1. The left flowers will be discarded. Of course, sometimes she will clean the vases. Because there are too many vases, she randomly choose to clean the vases numbered from A to B(A <= B). The flowers in the cleaned vases will be discarded.
Input
The first line contains an integer T, indicating the number of test cases.
For each test case, the first line contains two integers N(1 < N < 50001) and M(1 < M < 50001). N is the number of vases, and M is the operations of Alice. Each of the next M lines contains three integers. The first integer of one line is K(1 or 2). If K is 1, then two integers A and F follow. It means Alice receive F flowers and try to put a flower in the vase A first. If K is 2, then two integers A and B follow. It means the owner would like to clean the vases numbered from A to B(A <= B).
Output
For each operation of which K is 1, output the position of the vase in which Alice put the first flower and last one, separated by a blank. If she can not put any one, then output 'Can not put any one.'. For each operation of which K is 2, output the number of discarded flowers.
Output one blank line after each test case.
Sample Input
2 10 5 1 3 5 2 4 5 1 1 8 2 3 6 1 8 8 10 6 1 2 5 2 3 4 1 0 8 2 2 5 1 4 4 1 2 3
Sample Output
[pre]3 7 2 1 9 4 Can not put any one. 2 6 2 0 9 4 4 5 2 3 [/pre]
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题意：给你一个[0, N-1]的区间，区间的每个位置只能插一朵花。现在有Q次操作
1 x y 表示从第x个位置开始查y多花，若一朵花也插不上输出"Can not put any one."，反之输出插花的最左位置和最右位置。
2 x y 表示查询区间[x, y]有多少朵花，并清空所有的花。
思路：用线段树维护区间最左的空位置first、最右的空位置last、空位置总数sum，只需二分找到插花的最右位置lastp，在[x, lastp]区间上操作就可以了。二分的时候去掉多余的花就可以了。
主要是PushUp的操作，一开始没考虑儿子区间的first和last可能不存在，WA了一次。。。
AC代码：
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