CF52C.Circular RMQ
普及/提高-
通过率:0%
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题目描述
You are given circular array a0,a1,...,an−1 . There are two types of operations with it:
- inc(lf,rg,v) — this operation increases each element on the segment [lf,rg] (inclusively) by v ;
- rmq(lf,rg) — this operation returns minimal value on the segment [lf,rg] (inclusively).
Assume segments to be circular, so if n=5 and lf=3,rg=1 , it means the index sequence: 3,4,0,1 .
Write program to process given sequence of operations.
给你一个圆形数组 a0, a1, ..., an - 1 。可以对它进行两种操作:
inc(lf,rg,v) - 此操作会将数据段 [lf, rg] (包括)中的每个元素增加 v ;
rmq(lf,rg) - 此操作会将数据段 [lf, rg] (包括)中的每个元素返回最小值。- 此操作将返回数据段 [lf, rg] (包含)上的最小值。
假设线段是循环的,因此如果 n = 5 和 lf = 3, rg = 1 ,则表示索引序列: 3, 4, 0, 1 .
编写处理给定操作序列的程序。
输入格式
The first line contains integer n ( 1<=n<=200000 ). The next line contains initial state of the array: a0,a1,...,an−1 ( −106<=ai<=106 ), ai are integer. The third line contains integer m ( 0<=m<=200000 ), m — the number of operartons. Next m lines contain one operation each. If line contains two integer lf,rg ( 0<=lf,rg<=n−1 ) it means rmq operation, it contains three integers lf,rg,v ( 0<=lf,rg<=n−1;−106<=v<=106 ) — inc operation.
输入
第一行包含整数 n ( 1<=n<=200000 )。下一行包含数组的初始状态: a0,a1,...,an−1 ( −106<=ai<=106 ),ai 为整数。第三行包含整数 m ( 0<=m<=200000 ), m - 操作数。接下来的 m 行各含一个操作数。如果该行包含两个整数lf,rg ( 0<=lf,rg<=n−1 ) 表示 rmq 次运算,包含三个整数 lf,rg,v ( 0<=lf,rg<=n−1;−106<=v<=106 ) — inc 次运算。
输出格式
For each rmq operation write result for it. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).
输出
为每个 rmq 操作写入结果。请不要在 C++ 中使用 %lld 特指符来读写 64 位整数。最好使用 cout(也可以使用 %I64d)。
输入输出样例
输入#1
4 1 2 3 4 4 3 0 3 0 -1 0 1 2 1
输出#1
1 0 0