Robin Hood CodeForces - 672D (二分)

大意: 给定数组$a$, 每次操作使最大元素减小1最小元素增大1, 求k次操作后最大值与最小值的差.

二分出k次操作后最大值的最小值以及最小值的最大值, 若和能平分答案即为$max(0,R-L)$, 否则为$max(1,R-L)$

#include <iostream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head
 
#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif
 
int n, k, a[N];
int main() {
	scanf("%d%d", &n, &k);
	REP(i,1,n) a[i]=rd();
	int l=0,r=1e9,L,R;
	while (l<=r) {
		ll d = 0;
		REP(i,1,n) if (a[i]<mid) d+=mid-a[i];
		if (d<=k) L=mid,l=mid+1;
		else r=mid-1;
	}
	l=0,r=1e9;
	while (l<=r) {
		ll d = 0;
		REP(i,1,n) if (a[i]>mid) d+=a[i]-mid;
		if (d<=k) R=mid,r=mid-1;
		else l=mid+1;
	}
	int ans = max(0,R-L);
	ll sum = 0;
	REP(i,1,n) sum+=a[i];
	if (sum%n) ans=max(ans,1);
	printf("%d\n", ans);
}