이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//Nguyen Huu Hoang Minh
#include <bits/stdc++.h>
#define sz(x) int(x.size())
#define all(x) x.begin(),x.end()
#define reset(x) memset(x, 0,sizeof(x))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define N 2005
#define remain(x) if (x > MOD) x -= MOD
#define ii pair<int, int>
#define iiii pair< ii , ii >
#define viiii vector< iiii >
#define vi vector<int>
#define vii vector< ii >
#define bit(x, i) (((x) >> (i)) & 1)
#define Task "test"
#define int long long
using namespace std;
typedef long double ld;
const int inf = 1e10;
const int minf = -1e10;
int n;
double x[N], r[N];
double R[N];
void readfile()
{
ios_base::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
if (fopen(Task".inp","r"))
{
freopen(Task".inp","r",stdin);
//freopen(Task".out","w",stdout);
}
cin >> n;
for(int i=1; i<=n; i++){
cin >> x[i] >> r[i];
}
}
double check(int pos, int pre){
//(x[pre],R[pre])
//(x[pos],R[pos])
//dis*dis = (xpre-xpos)*(xpre-xpos) + (Rpre-R)*(Rpre-R) = Rpre^2 + 2*Rpre*R + R^2
//(xpre-xpos)*(xpre-xpos) = 4*Rpre*R
double R_cur = (x[pre]-x[pos])*(x[pre]-x[pos])/4/R[pre];
return R_cur;
}
void proc()
{
stack<pair<double,double>> st;
st.push(make_pair(x[1],r[1]));
for(int i=2; i<=n; i++){
while (st.size()){
double x1 = st.top().fi;
double r1 = st.top().se;
r[i] = min(r[i],0.25*(x1-x[i])*(x1-x[i])/r1);
if (r[i]>r1) st.pop();
else break;
}
st.push(make_pair(x[i],r[i]));
}
cout << setprecision(3) << fixed;
for(int i=1; i<=n; i++) cout<<r[i]<<'\n';
}
signed main()
{
readfile();
proc();
return 0;
}
컴파일 시 표준 에러 (stderr) 메시지
bal.cpp: In function 'void readfile()':
bal.cpp:37:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
37 | freopen(Task".inp","r",stdin);
| ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~
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