Slay the Dragon 二分 + 贪心-蒲公英云

Slay the Dragon 二分 + 贪心

二分+贪心找到最小花费，主要是边界有点麻烦

// Problem: C. Slay the Dragon
// Contest: Codeforces - Educational Codeforces Round 114 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1574/problem/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)
#include<iostream>
#include<cstdio>
#include<string>
#include<ctime>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<stack>
#include<climits>
#include<queue>
#include<map>
#include<set>
#include<sstream>
#include<cassert>
#include<bitset>
#include<list>
#include<unordered_map>
using namespace std;
#define ff first
#define ss second
#define lowbit(x) (x&-x)
#define pf(a) printf("%d\n",a)
#define mem(x,y) memset(x,y,sizeof(x))
#define dbg(x) cout << #x << " = " << x << endl
#define rep(i,l,r) for(int i = l; i <= r; i++)
#define fep(i,a,b) for(int i=b; i>=a; --i)
typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned long long ull;
const int N=2e5+100;
int n, m;
ll a[N];
vector<int> v[N];
void solve()
{ 
    scanf("%d", &n);
    ll sum = 0;
    a[0] = 1e18;
    rep(i,1,n) 
    { 
        scanf("%lld", &a[i]);
        sum += a[i];
    }    
    sort(a+1, a+1+n);
    scanf("%d", &m);
    while(m--)
    { 
        ll defen, atta, ans = 0;
        scanf("%lld %lld", &defen, &atta);
        int t = lower_bound(a+1, a+1+n, defen) - a;
        if(a[t] == defen)
        { 
            if(sum-a[t]>=atta) ans = 0;
            else ans = abs(a[t]+atta-sum);
        }
        else if(t > n) 
        { 
            ans += defen - a[n];
            if(sum - a[n] < atta) ans += (atta - sum + a[n]);
        }
        else
        { 
            ll res = 0;
            res += abs(a[t-1]-defen);
            ll s1 = sum - a[t-1];
            if(s1 < atta) res += atta - s1;
            ll xx = 0;
            if(a[t] >= defen) { 
                ll s3 = sum - a[t];
                if(s3 < atta) xx += atta - s3;
            }
            ll cnt = 0;
            cnt += abs(a[t]-defen);
            ll s2 = sum - a[t];
            if(s2 < atta) cnt += atta - s2;
            ans = min({ res, cnt,xx});
        }
        printf("%lld\n", ans);
        //cout << " t = " << t << " " << ans << endl;
        //cout << t << endl;
    }
}
int main()
{ 
        solve();
    return 0;
}