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); }