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题目链接:
Time Limit: 6000/3000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <bits/stdc++.h> #include <stack> #include <map> using namespace std; #define For(i,j,n) for(int i=j;i<=n;i++) #define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; template<class T> void read(T&num) { char CH; bool F=false; for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar()); for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar()); F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) { if(!p) { puts("0"); return; } while(p) stk[++ tp] = p%10, p/=10; while(tp) putchar(stk[tp--] + ‘0‘); putchar(‘\n‘); } const LL mod=1e9+7; const double PI=acos(-1.0); const LL inf=1e18; const int N=1e5+10; const int maxn=5e3+4; const double eps=1e-12; double v[N],pos[N],t,c; int d[N],k,n; int main() { int T; read(T); while(T--) { read(n); scanf("%lf",&c); For(i,1,n) { scanf("%lf%lf%d",&v[i],&pos[i],&d[i]); } sort(v+1,v+n+1); int q; read(q); while(q--) { scanf("%lf%d",&t,&k); printf("%.3lf\n",sqrt(v[k]*v[k]+2*c*t)); } } return 0; }
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原文地址:http://www.cnblogs.com/zhangchengc919/p/5762183.html