https://ac.nowcoder.com/acm/contest/331/A
C++版本一
题解:
std
如果只有一根木条,显然答案就是一个圆弧。
当有两根木条的时候,问题等价于在这个圆弧上任一点放置木条2。
显然可以发现可以到达的位置是一个圆环或者一个圆(当且仅当l1=l2)。
#include <bits/stdc++.h>
using namespace std;
int l1, l2;
int main() {
int T;
scanf("%d%d%d", &l1, &l2, &T);
while (T--) {
double x, y;
scanf("%lf%lf", &x, &y);
long double dist = hypotl(x, y);
long double ans = 0;
if (dist > l1 + l2) ans = dist - l1 - l2;
if (dist < abs(l1 - l2)) ans = abs(l1 - l2) - dist;
printf("%.12Lf\n", ans);
}
return 0;
}
C++版本二
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-7;
const int INF = 0x3f3f3f3f;
int t,n,m,k,q;
double x,y;
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
while(~scanf("%d%d",&n,&m)){
scanf("%d",&t);
if(n<m)
swap(n,m);
while(t--){
scanf("%lf%lf",&x,&y);
double ans=sqrt(x*x+y*y);
if((n-m<=ans&&ans<=n+m)){
cout<<0<<endl;
}else{
if(ans<n-m)
printf("%.8lf\n",n-m-ans);
else
printf("%.8lf\n",ans-n-m);
}
}
}
//cout << "Hello world!" << endl;
return 0;
}