ICPC World Finals 2018 Problem H Single Cut of Failure

题目链接
 题解视频
 题解文档

解法概要：
问题可以转化为

考虑一个长为 $2n$ 的数组 $A$，$1$ 到 $n$ 这 $n$ 个整数每个恰在 $A$ 中出现 $2$ 次。判断是否存在一个长为 $n$ 的子段使得 $1$ 到 $n$ 在其中各出现一次。

这个经典问题可用双指针在 $O(n)$ 的时间内解决。

这道题我 WA 到死，不是因为思路有漏洞，而是因为我不了解 std::cout 。

什么是 IO manipulators

默认情况下，cout 输出浮点数的格式是 std::defaultfloat。

下面我们来看一下 defaultfloat 究竟是个什么东西。

defaultfloat 是一个 ios_base manipulator 。

那么 manipulator 是什么呢

IO manipulator are helper functions that make it possible to control input/output streams using operator<< or operator>>.

The manipulators that are invoked without arguments (e.g. std::cout << std::boolalpha; or std::cin >> std::hex;) are implemented as functions that take a reference to a stream as their only argument.（e.g. std::fixed is declared as ios_base& fixed(ios_base& str);）The special overloads of basic_ostream::operator<< and basic_istream::operator>> accept pointers to these functions.

The manipulators that are invoked with arguments (e.g. std::cout << std::setw(10);) are implemented as functions returning objects of unspecified type. These manipulators define their own operator<< or operator>> which perform the requested manipulation. Manipulators that take arguments are defined in the iomanip header.
SOURCE

`basic_ostream::operator<<` 默认情况下如何输出浮点数

By default, floating-point values are printed using six digits of precision (1.55 printed as 1.55, 1.55667 printed as 1.55667, 3.1415926 printed as 3.14159); the decimal point is omitted if the value has no fractional part (e.g. 1.00 is printed as 1); trailing zeros are removed (e.g. 1.5000 printed as 1.5); and they are printed in eighter fixed or scientific notation depending on the value of the number. The library chooses a format that enhances readability of the number. Very large and very small values are printed using scientific notation. Other values are printed in fixed decimal.

总结：
默认情况下（对应于 manipulator std::defaultfloat），basic_ostream::operator<< 输出浮点数的格式为：
最多 6 位数字（整数部分和小数部分加起来共 6 位，术语为 precision 为 6），具体而言

若原本整数部分与小数部分位数之和就不超过 6 位，则原样输出，并去除 trailing zeros。
若整数部分不超过 6 位，则先四舍五入去除多出小数部分再去除 trailing zeros。
在上述两种情况中，若小数部分完全被去除，则小数点也不输出。
若整数部分超过 6 位，则用科学记数法。

`std::setprecision(n)` 与 `std::fixed` 的作用

setprecision(n) 的作用是将浮点数的输出精度 (precision) 设为 n。那么这里的 precision 究竟是什么意思呢？

By default, precision controls the total number of digits that are printed. When printed, floating-point values are rounded, not truncated, to the current precision. Thus, if the current precision is four, then 3.14159 becomes 3.142; if the precision is three, then it is printed as 3.14.

We can change precision by calling the precision by calling the precision member (function) or by using the setprecision manipulator. The precision member is overloaded. One version takes an int value and sets the precision to that new value. It returns the previous value. The other version takes no arguments and returns the current precision value. The setprecison manipulator takes an argument, which it uses to set the precision.

C++ Primer, 5th edition, pp 756

注意，这里所说的 “rounded to the current precision” 具体仍然是按上面总结的方法操作。

std::fixed 控制浮点数的记数方式（不是计数方式，也不是精度），std::fixed 使得浮点数始终按十进制格式（fixed decimal，不理解 fixed 在这里是什么意思；fixed decimal 与 scientific notation 相对）输出，也就是说较大的数或较小的数不用科学记数法了。

另外使用 fixed 这个 manipulator 之后，精度的含义也改变了。

These manipulators (fixed, scientific, hexfloat) also change the default meaning of the precision for the stream. After executing scientific, fixed, or hexfloat, the precision value controls the number of digits after the decimal point. By default, precision specifies the total number of digits——both before and after the decimal point.

总结

如果题目中要输出浮点数，若指明保留几位小数（例如 4 位小数），则在 main 函数的开头加上一行

std::cout << std::fixed << std::setprecision(4);

若只写

std::cout << std::fixed;

就是保留 6 位小数。

Reference

https://www.uow.edu.au/~lukes/TEXTBOOK/notes-cpp/io/omanipulators.html