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转载自：http://www.cnblogs.com/think/archive/2011/10/06/2199692.html Subversion properties 在CEF开发中，应当如下将Subversion配置文件配置成自动设置新文件的属性，不要使用svn:eol-style=native因为它会使得不同平台间的文件比较变得十分痛苦 # CEF-specific config file to put at ~/.subversion/config or %USERPROFILE%\AppData\Roaming\Subversion\config# Originally copied from http://src.chromium.org/viewvc/chrome/trunk/tools/buildbot/slave/config?revision=46073[miscellany]global-ignores = *.pyc *.user *.suo *.bak *~ #*# *.ncb *.o *.lo *.la .*~ .#* .DS_Store .*.swp *.scons *.mk *.Makefile *.sln *.vcproj *.rules SConstruct *.xcodeprojenable-auto-props = yes[auto

link #include <cstdio> #include <cmath> #include <vector> #include <algorithm> #include <climits> #include <unordered_map> #include <cstdio> #include <iostream> # define LL long long using namespace std; struct Station{ double dis; double price; }; int main(){ double cmax, D, davg; int N; scanf("%lf %lf %lf %d", &cmax, &D, &davg, &N); vector<Station> stas(N+1); stas[0].dis=D; stas[0].price=0; for(int i=1;i<=N;i++){ scanf("%lf %lf", &stas[i].price, &stas[i].dis); } sort(stas.begin(),stas.end(),[](Station s1, Station s2){ return s1.dis<s2.dis; }); if(stas[0].dis!=0){ printf("The maximum travel

#include <stdio.h> #include <stdlib.h> #define PI 3.14159 int main() { double R,l,s; scanf("%lf",&R); l = 2 * PI * R; s = PI * R * R; printf(“圆的周长为：%lf\n”,l); printf(“圆的面积为：%lf”,s); system(“pause”); return 0 ; } 来源： CSDN 作者： K2MnO2 链接： https://blog.csdn.net/Polary1/article/details/104560671

A欧几里得 1 0 2 1 3 2 第一个数字是上一行数字的和，第二个数字是上一行数字第一个数字 #include <cstdio> #include <cmath> #include <algorithm> #include <iostream> #include <cstring> #include <queue> #include <set> #include <map> #include <stack> using namespace std; #define scd(a) scanf("%d",&a) #define scdd(a,b) scanf("%d%d",&a,&b) #define scddd(a,b,c) scanf("%d%d%d",&a,&b,&c) #define scl(a) scanf("%lld",&a) #define scll(a,b) scanf("%lld%lld",&a,&b) #define sclll(a,b,c) scanf("%lld%lld%lld",&a,&b,&c) #define prl(a) printf("%lld\n",a) #define prd(a) printf("%d\n",a) #define prf(a) printf("%lf\n",a) #define ptd(a) printf("%d ",a)

#include <stdio.h> #define pi 3.1415926 int main() { double a; scanf("%lf", &a); printf("正方形面积:%lf\n", a * a); printf("内接圆面积:%lf\n", pi * a * a / 4); printf("外接圆面积:%lf\n", pi * a * a / 2); return 0; } 来源： https://www.cnblogs.com/bobotongxue/p/12290647.html

12点四十起床饭都没恰，温州阴郁的天气隐隐暗示了今天的基调 https://ac.nowcoder.com/acm/contest/3003#question A.做游戏 石头剪刀布wa两发娱乐一下，int的read加起来的时候在右边炸了一回，重新写sclll没抄对，甚至样例都没过就直接上了，这脑残的行径能怪谁呢 #include <cstdio> #include <cmath> #include <algorithm> #include <iostream> #include <cstring> #include <queue> #include <set> #include <map> #include <stack> using namespace std; #define scd(a) scanf("%d",&a) #define scdd(a,b) scanf("%d%d",&a,&b) #define scddd(a,b,c) scanf("%d%d%d",&a,&b,&c) #define scl(a) scanf("%lld",&a) #define scll(a,b) scanf("%lld%lld",&a,&b) #define sclll(a,b,c) scanf("%lld%lld%lld",&a,&b,&c) #define prl(a) printf(

lf rt 确定子串范围 ， mm记录曾经遇到过的最长子串 针对每一个新的字符，从右向左找是否有重复，重复则重新划定范围，舍弃先遇到的字符。 class Solution { public : int lf = 0 ; int rt = 0 ; int lengthOfLongestSubstring ( string s ) { if ( s . length ( ) == 1 ) return 1 ; int mm = 0 ; for ( int i = 1 ; i < s . length ( ) ; i ++ ) { for ( int j = rt ; j >= lf ; j -- ) if ( s [ i ] == s [ j ] ) lf = j + 1 ; rt ++ ; mm = mm > ( rt - lf + 1 ) ? mm : rt - lf + 1 ; } return mm ; } } ; 来源： CSDN 作者： 啾九啾 链接： https://blog.csdn.net/qq_40070622/article/details/104095630

传送门 思路 ：这是一道推式子题 假设人离灯泡的地面距离为x 那么影子长D+H−(x+(H−h)∗D/x)① 定义域为 x<=D且x>=D - h * D / H 应用 均值不等式 原式<=D+H-2 sqrt(D (H-h)) 当x=sqrt(D*(H-h))时取等号 设a=sqrt(D*(H-h)) ①式图像为单峰函数，到分类讨论的时间了。如果a>D，最大值在x=D时取到，此时最大值为h;如果a<D-h D/H,最大值在x=D-h D/H时取到，此时最大值为h D/H;其他情况的最大值在x=a是取到，最大值为D+H-2 a 代码如下： # include <cmath> # include <cstring> # include <cstdio> # include <algorithm> # include <queue> # include <fstream> # include <stack> # include <iostream> # include <time.h> # define LL long long using namespace std ; LL T ; double H , h , D ; int main ( ) { cin >> T ; while ( T -- ) { scanf ( "%lf%lf%lf" , & H , & h , & D )

#include "stdio.h" #include "math.h" void main() { double a,b,c,x1,x2,p,q,dis; printf("输入abc的值\n"); scanf("a=%lf b=%lf c=%lf",&a,&b,&c); dis=sqrt(b*b-4*a*c); p=-b/(2*a); q=dis/(2*a); x1=p+q; x2=p-q; printf("一元二次方程的解为：\n"); printf("x1=%5.2lf\nx2=%5.2lf\n",x1,x2); } 来源： https://www.cnblogs.com/zyz322/p/12232466.html