标签:
2
5
2 3.2 4 4.5 6
10
1 2 3 1 2 1.2 3 1.1 1 2
2
5
分析:依次选取半径最大的雷达即可;
代码:
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <map> #include <queue> #include <stack> #include <vector> #include <list> #include <ext/rope> #define rep(i,m,n) for(i=m;i<=n;i++) #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++) #define vi vector<int> #define pii pair<int,int> #define mod 1000000007 #define inf 0x3f3f3f3f #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) const int maxn=1e3+10; const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}}; using namespace std; using namespace __gnu_cxx; ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;} int n,m,cnt; double a[maxn],now; int main() { int i,j,k,t; scanf("%d",&t); while(t--) { scanf("%d",&n); rep(i,0,n-1)scanf("%lf",&a[i]); sort(a,a+n); cnt=0;now=0; for(i=n-1;i>=0;i--) { if(now>=20)break; now+=2*sqrt(a[i]*a[i]-1); cnt++; } printf("%d\n",cnt); } //system ("pause"); return 0; }
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原文地址:http://www.cnblogs.com/dyzll/p/5697007.html