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iomanip
hdu1053 Entropy
1 #include<iostream> 2 #include<
iomanip
> 3 #define N 27 4 using namespace std; 5
·
2015-11-08 14:31
HDU
最长公共子序列的C++实现---附二维指针的使用方法
#include<iostream> #include<
iomanip
> using namespace std; //tag标志,0为左斜上,1取左,2取上;count为最长公共子序列计数
·
2015-11-08 14:15
C++
C++ double 小数精度控制
fixed<<setprecision(20)<<mydouble<<endl; #include <iostream> #include <
iomanip
·
2015-11-08 10:17
double
[原]C++ double 小数精度控制
fixed<<setprecision(20)<<mydouble<<endl; #include <iostream> #include <
iomanip
·
2015-11-08 10:16
double
四则运算的扩展
2 #include<conio.h> 3 #include<cstdlib> 4 #include<ctime> 5 #include <
iomanip
·
2015-11-08 10:21
四则运算
C++ cout 输出 16, 8 , 2进制
(转载) #include <iostream> #include <
iomanip
> #include <bitset> using std::bitset
·
2015-11-08 09:54
C++
C++基础——格式化输出
准备头文件包含
iomanip
头文件,
iomanip
顾名思义,
iomanip
u
lanchunhui
·
2015-11-07 18:00
C++
格式化输出
iomanip
Print a rhombus (C++)
一个内置方法轻描淡写地就把问题给解决了: cout<<setw(n)<<n<<endl; 打印n个空格(为了辨别,后面加了一个字母n)不过要加#include <
iomanip
·
2015-11-07 15:21
print
GDAL读取tiff文件/C++源码
include "stdafx.h" #include "gdal_priv.h" #include <iostream> #include <
iomanip
·
2015-11-07 15:35
C++
hdu 4430 Yukari's Birthday
#include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 14:38
birt
hdu 4578 Transformation 线段树
include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 14:38
transform
hdu 1536/1944 / POJ 2960 / ZOJ 3084 S-Nim 博弈论
#include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 14:31
poj
hdu 4417 Super Mario 离线线段树
#include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 14:22
super
hdu 4412 Sky Soldiers DP
#include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 14:22
HDU
hdu 4638 Group
#include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 14:17
group
poj 1026 Cipher
#include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 14:13
poj
hdu 1080 Human Gene Functions
#include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 14:12
functions
poj 1850 Code
#include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 14:12
code
2013 Multi-University Training Contest 2 Balls Rearrangement
#include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 14:09
test
hdu 1521 排列组合
#include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
>#include
·
2015-11-07 14:06
HDU
2013 Multi-University Training Contest 1 Partition
这题主要是推公式…… #include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
·
2015-11-07 14:04
partition
2013 Multi-University Training Contest 1 I-number
水题,注意不要去掉前导0…… #include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
·
2015-11-07 14:04
number
hdu 1905 小数化分数2
#include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
>#include
·
2015-11-07 14:03
HDU
hdu 1755 A Number Puzzle
这题枚举k节省时间 #include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
·
2015-11-07 14:02
number
hdu 1796 How many integers can you find
#include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 14:02
Integer
hdu 2837 Calculation
#include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 14:01
HDU
hdu 3758 Factorial Simplification
include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 14:59
imp
hdu 4187 Alphabet Soup
6866991 #include<iostream> #include<stdio.h> #include<algorithm> #include<
iomanip
·
2015-11-07 14:58
Alpha
hdu 3123 GCC
#include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
>#include
·
2015-11-07 14:56
gcc
hdu 3923 Invoker
完全是套用polya模版…… #include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
·
2015-11-07 14:53
HDU
hdu 1352 I Conduit!
主要是排序问题,其他都很好做…… #include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
·
2015-11-07 14:52
HDU
2013 ACM-ICPC长沙赛区全国邀请赛——Bottles Arrangement
+m-1+m-1+……前n项 #include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
·
2015-11-07 14:51
ICPC
2013 ACM-ICPC长沙赛区全国邀请赛—Special equations
直接跳过了 这题直接枚举就可以过了 #include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
·
2015-11-07 14:51
ICPC
2013 ACM-ICPC长沙赛区全国邀请赛——A So Easy!
#include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 14:50
ICPC
hdu 1376 Octal Fractions
)[10],从低位向高位计算 #include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
·
2015-11-07 14:49
action
hdu 1713 相遇周期
#include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
>#include
·
2015-11-07 14:49
HDU
hdu 1329 Hanoi Tower Troubles Again!
1+(i+i%2)/2*2; #include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
·
2015-11-07 14:48
HDU
hdu 1333 Smith Numbers
刚开始没看清题意,要找的数一定要是素数 #include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
·
2015-11-07 14:47
number
hdu 1309 Loansome Car Buyer
纯粹的阅读理解题………… #include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
·
2015-11-07 14:47
HDU
hdu 1288 Hat's Tea
这个要慢慢理解…… #include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
·
2015-11-07 14:46
HDU
hdu 1284 钱币兑换问题
简单水题…… #include<iostream>#include<stdio.h>#include<algorithm>#include<
iomanip
>
·
2015-11-07 14:45
HDU
poj 3292 Semi-prime H-numbers
3 #include<algorithm> 4 #include<vector> 5 #include<cmath> 6 #include<
iomanip
·
2015-11-07 14:38
number
hdu 3579 Hello Kiki
include<iostream>#include<stdio.h>#include<algorithm>#include<cmath>#include<
iomanip
·
2015-11-07 14:37
HDU
hdu 1134 Game of Connections
include<iostream>#include<stdio.h>#include<algorithm>#include<cmath>#include<
iomanip
·
2015-11-07 14:35
Connection
HDOJ 1757 A Simple Math Problem
矩阵快速幂的水题 #include<iostream>#include<stdio.h>#include<cmath>#include<
iomanip
>
·
2015-11-07 14:26
simple
poj 1528 Perfection
代码如下: #include<iostream> #include<
iomanip
> using namespace std; int an[60001]; int
·
2015-11-07 14:23
poj
牛顿迭代法 POJ 2868
– (x – A/x) / 2 根据这条公式迭代就可以 终止条件是前后两次迭代出来的值少于1e-6 #include <iostream> #include <
iomanip
·
2015-11-07 13:58
poj
hdu 4753 Fishhead’s Little Game 博弈论+记忆化搜索
#include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 10:27
game
hdu 4352 XHXJ's LIS 数位DP
#include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 10:22
HDU
hdu 4745 Two Rabbits
#include<iostream> 2 #include<stdio.h> 3 #include<algorithm> 4 #include<
iomanip
·
2015-11-07 10:20
HDU
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