AC了，PI要精确到小数点后7位……

只差一个点，没AC

#include<iostream>
#include<cmath>
#include<cstdio>
float x[101],y[101],r;
using namespace std;
int main()
{
  int n;
  cin>>n>>r;
  for(int i=0;i<n;i++) cin>>x[i]>>y[i];
  x[n]=x[0];
  y[n]=y[0];
  float total=0;
  for(int i=0;i<n;i++){
  total+=sqrt((x[i]-x[i+1])*(x[i]-x[i+1])+(y[i]-y[i+1])*(y[i]-y[i+1]));
  }
  total+=3.14*r*2;
  printf("%.2f",total);
  return 0;
}

谢了，我发给你的东西你收到没？

全AC了
```c++
评测结果
编译成功

测试数据 #0: Accepted, time = 15 ms, mem = 556 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 552 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 552 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 556 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 556 KiB, score = 10
Accepted, time = 15 ms, mem = 556 KiB, score = 60
代码
#include<iostream>
#include<cmath>
#include<cstdio>
float x[101],y[101],r;
using namespace std;
int main()
{
int n;
cin>>n>>r;
for(int i=0;i<n;i++) cin>>x[i]>>y[i];
x[n]=x[0];
y[n]=y[0];
float total=0;
for(int i=0;i<n;i++){
total+=sqrt((x[i]-x[i+1])*(x[i]-x[i+1])+(y[i]-y[i+1])*(y[i]-y[i+1]));
}
total+=3.1415926*r*2;
printf("%.2f",total);
return 0;
}

因为你数组下标使用0~n-1
所以
x[n+1]=x[0];
y[n+1]=y[0];
要改成
x[n]=x[0];
y[n]=y[0];

只AC三个点

c++
#include<iostream>
#include<cmath>
#include<cstdio>
float x[101],y[101],r;
using namespace std;
int main()
{
  int n;
  cin>>n>>r;
  for(int i=0;i<n;i++) cin>>x[i]>>y[i];
  x[n+1]=x[0];
  y[n+1]=y[0];
  float total=0;
  for(int i=0;i<n;i++){
  total+=sqrt((x[i]-x[i+1])*(x[i]-x[i+1])+(y[i]-y[i+1])*(y[i]-y[i+1]));
  }
  total+=3.14*r*2;
  printf("%.2f",total);
  return 0;
}

#include <cstdio>
#include <cmath>
using namespace std;
int main() {
  int n;
  float a[101][3],ans = 0,r;
  scanf("%d %f\n",&n,&r);
  for (int i = 1;i <= n;i++)
    scanf("%f %f",&a[i][1],&a[i][2]);
  for(int i = 1;i < n;i++)
    ans += sqrt((a[i][1]-a[i+1][1])*(a[i][1]-a[i+1][1])+(a[i][2]-a[i+1][2])*(a[i][2]-a[i+1][2]));
  ans += sqrt((a[n][1]-a[1][1])*(a[n][1]-a[1][1])+(a[n][2]-a[1][2])*(a[n][2]-a[1][2]));
  printf("%.2f",ans+r*2*3.1415926);
  return 0;
}

一个数，绳子的长度，**精确到小数点后2位**。