#include<cstdio>
#include<cmath>

using namespace std;

int main()
{
int N;
double R,x[101],y[101];
scanf("%d%lf",&N,&R);
for(int i=1;i<=N;i++)
scanf("%lf%lf",x+i,y+i);
x[N+1]=x[1];
y[N+1]=y[1];
double ans=3.141592*R*2.0;
for(int i=1;i<=N;i++)
ans+=sqrt((x[i+1]-x[i])*(x[i+1]-x[i])+(y[i+1]-y[i])*(y[i+1]-y[i]));
printf("%.2lf",ans);
return 0;
}

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main()
{
int N,n;
float R,X2,Y2,X1,Y1,dis=0,sum=0,s1,s2;
scanf("%d %f",&N,&R);
if(N==1)
{scanf("%f %f",&X2,&Y2);
printf("%.2f",2*R*3.141);}
else
{scanf("%f %f",&X2,&Y2);
s1=X2,s2=Y2;
for(n=2;n<=N;n++)
{X1=X2;Y1=Y2;
scanf("%f %f",&X2,&Y2);
dis=sqrt((X1-X2)*(X1-X2)+(Y1-Y2)*(Y1-Y2));
sum=sum+dis;
}
printf("%.2f",sum+2*R*3.141+sqrt((s1-X2)*(s1-X2)+(s2-Y2)*(s2-Y2)));}

return 0;
}

#include<cstdio>
#include<cstdlib>
#include<cmath>
int main()
{
double pi=3.1415926,len=0,c,r;
double x[10],y[10];
int n,i;
// freopen("in.in","r",stdin);
// freopen("out.out","w",stdout);
scanf("%d%lf",&n,&r);
for(i=1;i<=n;i++)
scanf("%lf%lf",&x[i],&y[i]);
x[n+1]=x[1];
y[n+1]=y[1];
for(i=1;i<=n;i++)
len+=sqrt(pow(x[i]-x[i+1],2)+pow(y[i]-y[i+1],2));
c=r*pi*2;
len+=c;
printf("%.2lf",len);
// fclose(stdin);
//fclose(stdout);
//system("pause");
return 0;
}

为什么，这段代码回运行错误

测试数据 #0: Accepted, time = 0 ms, mem = 248 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 248 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 244 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 248 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 248 KiB, score = 10
测试数据 #5: Accepted, time = 15 ms, mem = 244 KiB, score = 10
做水题加信心2333

include <stdio.h>

include <math.h>

double q(double a,double b,double c,double d)
{
return sqrt((c-a)*(c-a)+(d-b)*(d-b));}
int main()
{
int i,N,j;
double a,b,c,d,r,s,k,l;
const double pi = 4.0 * atan(1.0);
scanf("%d%lf", &N,&r);
s = 2 * pi * r;
scanf("%lf%lf", &a, &b);
for(i = 2;i <= N;i++)
{
if(i % 2 == 0) scanf("%lf%lf", &c, &d);
else scanf("%lf%lf", &a, &b);
s += q(a,b,c,d);
if(i == 2) {k = a;l = b;}
if(i == N) s += (i % 2 == 0) ? q(c,d,k,l) : q(a,b,k,l);}
printf("%.2lf\n", s);
}

水………………

记录信息
评测状态 Accepted
题目 P1007 绕钉子的长绳子
递交时间 2014-08-04 09:39:04
代码语言 C++
评测机 VijosEx
消耗时间 30 ms
消耗内存 272 KiB
评测时间 2014-08-04 09:39:09
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 268 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 272 KiB, score = 10
测试数据 #4: Accepted, time = 15 ms, mem = 272 KiB, score = 10
测试数据 #5: Accepted, time = 15 ms, mem = 272 KiB, score = 10
Accepted, time = 30 ms, mem = 272 KiB, score = 60

#include <cstdio>
#include <cmath>
const double Pi = acos(-1.0);
int main()
{ int n;
double r;
scanf("%d%lf", &n, &r);
double x[n], y[n];
for (int i = 0; i < n; i++)
scanf("%lf%lf", &x[i], &y[i]);
double ans = 2*Pi*r;
for (int i = 0; i < n; i++)
ans += sqrt((x[i]-x[(i+1)%n])*(x[i]-x[(i+1)%n])+(y[i]-y[(i+1)%n])*(y[i]-y[(i+1)%n]));
printf("%.2lf\n", ans);
return 0;
}

算法：
题目已经说了是逆时针了， 所以只要s:=s+len(a[i].x,a[i].y,a[i-1].x,a[i-1].y);
整个绳子绕在圆上的长度合起来是一个整圆

Vijos 题解：http://hi.baidu.com/umule/item/2c997f8ed9600fdae596e017
有疑问请留言 共同进步

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define PI 3.1415926
#define REP( i, n ) for ( int i = 1; i <= n; i ++ )
#define FOR( i, a, b ) for ( int i = a; i <= b; i ++ )
#define DWN( i, a, b ) for ( int i = b; i >= a; i -- )
#define RST( a, x ) memset ( a, x, sizeof ( a ) );
#define ONLINE_JUDGE NULL

using namespace std;
double x[ 200 ], y[ 200 ], r, ans, t1, t2;
int n;
int main ()
{

#ifndef ONLINE_JUDGE
freopen ( "P1000.in", "r", stdin );
freopen ( "P1000.out", "w", stdout );
#endif

scanf ( "%d%lf", &n, &r );
ans = r * 2 * PI;
REP ( i, n )
{
scanf ( "%lf%lf", &x[ i ], &y[ i ] );
if ( i > 1 ) t1 = x[ i ] - x[ i - 1 ], t2 = y[ i ] - y[ i - 1 ];
ans += sqrt ( t1 * t1 + t2 * t2 );
}
t1 = x[ 1 ] - x[ n ], t2 = y[ 1 ] - y[ n ];
ans += sqrt ( t1 * t1 + t2 * t2 );
printf ( "%.2f\n", ans );
return 0;
}
其实这题很弱菜……

一上来把R定义成int了。。。神坑我三次WA。
简单题，一上来还以为好难，用解析几何做了个算法，做了一半才发现自己傻X了，明明是凸多边形啊。。。线绕过的正好是一个圆啊。。。
23行AC
C++ Code:
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
int main(){
int n;
double r;
cin>>n;
cin>>r;
double p[101][2];
double ans=0;
for(int i=0;i<n;i++){
cin>>p[i][0];
cin>>p[i][1];
}
p[n][0]=p[0][0];
p[n][1]=p[0][1];
for(int i=1;i<=n;i++){
ans+=hypot(fabs(p[i][0]-p[i-1][0]),fabs(p[i][1]-p[i-1][1]));
}
ans+=2.00000000*3.14159265358*r;
cout<<setprecision(2)<<std::fixed<<ans;
}

不想说了 已经是凸包了 没看到 还写了个找凸包的....... 心都碎了

一开始很烦啊，后来想通就好了

题解

var
--n,i:longint;
--a:array[1..100,1..2]of extended;
--long,x,r:extended;
begin
--read(n,r);
--for i:=1 to n do readln(a[i,1],a[i,2]);
--for i:=1 to n-1 do
--begin
---x:=sqrt(sqr(abs(a[i,1]-a[i+1,1]))+sqr(abs(a[i,2]-a[i+1,2])));
---long:=long+x;
--end;
--x:=sqrt(sqr(abs(a[n,1]-a[1,1]))+sqr(abs(a[n,2]-a[1,2])));
--long:=long+x+2*pi*r;
--write(long:0:2);
end.

水题。。注意pi，abs的返回值即可。纯C++版本。有钉子。

Block Code

#include<iostream>
#include<cmath>
#include<iomanip>
#define PI 3.1415926
using namespace std;

int main()
{
short n;double r;
cin>>n>>r;
double ans=PI*2.0*r;
double x,y,x1,y1,xs,ys;
cin>>x>>y;
xs=x;ys=y;
for(int i=2;i<=n;++i)
{
cin>>x1>>y1;
ans+=hypot(fabs(x1-x),fabs(y1-y));
x=x1;y=y1；
//cout<<ans<<"//";
}
ans+=hypot(fabs(xs-x),fabs(ys-y));
cout<<setiosflag(ios::fixed)<<setprecision(2)<<ans;
return 0;
}

题解:http://blog.163.com/syf_light_feather/blog/static/22375506720136288215949/?latestBlog

表示不能容忍半径是实型、坑了后三个点……

点击查看程序源码+详细题解

编译通过...

├ 测试数据 01：答案正确... (7ms, 200KB)

├ 测试数据 02：答案正确... (0ms, 200KB)

├ 测试数据 03：答案正确... (0ms, 200KB)

├ 测试数据 04：答案正确... (0ms, 200KB)

├ 测试数据 05：答案正确... (0ms, 200KB)

├ 测试数据 06：答案正确... (0ms, 200KB)

---|---|---|---|---|---|---|---|-

Accepted / 100 / 7ms / 200KB

P转C第二题

注意不能用3.14哦~我就错了一组！！