Submission #322152

#	Time	Username	Problem	Language	Result	Execution time	Memory
322152		zaneyu	Sure Bet (CEOI17_sure)	C++14	0 / 100	1 ms	364 KiB

sure

/*input
4
1.4 3.7
1.2 2
1.6 1.4
1.9 1.5
*/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<long long, null_type, less_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
typedef long long int ll;
#define ld double
#define pii pair<int,int>
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(ll i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
const ll maxn=1e5+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=(1e9+7);
const ll MOD2=1000002173;
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
ll mult(ll a,ll b){
    return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b,ll MOD){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=(res*a)%MOD;
        a=(a*a)%MOD;
        b>>=1;
    }
    return res;
}
ld a[maxn],b[maxn];
int main(){
	ios::sync_with_stdio(false),cin.tie(0);
	int n;
	cin>>n;
	REP1(i,n) cin>>a[i]>>b[i];
	sort(a+1,a+n+1);
	reverse(a+1,a+n+1);
	sort(b+1,b+n+1);
	reverse(b+1,b+n+1);
	REP1(i,n) a[i]+=a[i-1],b[i]+=b[i-1];
	ld ans=0;
	REP1(i,n){
		int p=lower_bound(b+1,b+n+1,a[i])-b;
		if(p>=1 and p<=n) MXTO(ans,a[i]-i-p);
		p=lower_bound(a+1,a+n+1,b[i])-a;
		if(p>=1 and p<=n) MXTO(ans,b[i]-i-p);
	}
	cout<<ans;
}

• Subtask 1: 0 / 20
• Subtask 2: 0 / 40
• Subtask 3: 0 / 40