http://poj.org/problem?id=1905
题意
一根两端固定在两面墙上的杆,受热后变弯曲。求前后两个状态的杆的中点位置的距离
分析
很明显需要推推公式。
由②的限制条件来二分角度,答案由①给出。注意,这种写法的精度要求较高。
#include<iostream> #include<cmath> #include<cstring> #include<queue> #include<vector> #include<cstdio> #include<algorithm> #include<map> #include<set> #define rep(i,e) for(int i=0;i<(e);i++) #define rep1(i,e) for(int i=1;i<=(e);i++) #define repx(i,x,e) for(int i=(x);i<=(e);i++) #define X first #define Y second #define PB push_back #define MP make_pair #define mset(var,val) memset(var,val,sizeof(var)) #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 pd(a) printf("%d\n",a) #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 IOS ios::sync_with_stdio(false);cin.tie(0) #define lc idx<<1 #define rc idx<<1|1 #define rson mid+1,r,rc #define lson l,mid,lc using namespace std; typedef long long ll; template <class T> void test(T a){cout<<a<<endl;} template <class T,class T2> void test(T a,T2 b){cout<<a<<" "<<b<<endl;} template <class T,class T2,class T3> void test(T a,T2 b,T3 c){cout<<a<<" "<<b<<" "<<c<<endl;} const int inf = 0x3f3f3f3f; const ll INF = 0x3f3f3f3f3f3f3f3fll; const ll mod = 1e9+7; int T; void testcase(){ printf("Case %d: ",++T); } const int MAXN = 1e5+10; const int MAXM = 30; int main() { #ifdef LOCAL freopen("data.in","r",stdin); #endif // LOCAL double l,n,c; while(~scanf("%lf%lf%lf",&l,&n,&c)){ if(l<0&&n<0&&c<0) break; double low=0.0,high=asin(1.0),mid; double L=(1+n*c)*l; while(high-low>1e-12){ mid=(high+low)/2; if(L*sin(mid)<=l*mid) high=mid; else low=mid; } printf("%.3f\n",l/2*tan(low/2)); } return 0; }