Codeforces #352 Div2 D Robin Hood(二分查找)

题目链接:

Codeforces #352 Div2 D Robin Hood

题意:

给出n个数,每次可以将最大数-1,最小数+1,问K次后最大数和最小数之差?

分析:

二分查找k次最大数的最小值y和最小数的最大值x.

如果x>=y并且这些数的和sum能被n整除则ans为0;

如果x>=y并且这些数的和sum不能被n整除则ans为1.

否则(x < y),ans为y-x.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <climits>
#include <cmath>
#include <ctime>
#include <cassert>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
using namespace std;
typedef long long ll;
const int MAX_N = 500010;

int n, k;
ll data[MAX_N];

bool check1(ll y)
{
ll kk = k;
for(int i = n - 1; i >= 0; i--){
if(data[i] <= y ) return true;
kk -= (data[i] - y);
if(kk < 0) return false;
}
return true;
}

ll binsearch1() //最大值的最小值
{
ll high = data[n - 1], low = data[0];
while(high > low){
ll mid = (high + low) >> 1;
if(check1(mid)) high = mid;
else low = mid + 1;
//cout << "high = " << high << " low = " << low << endl;
}
return high;
}

bool check2(ll x)
{
ll kk = k;
for(int i = 0; i < n; i++){
if(data[i] >= x) return true;
kk -= (x - data[i]);
if(kk < 0) return false;
}
return true;
}

ll binsearch2() //最小值的最大值
{
ll high = data[n - 1], low = data[0];
while(high > low + 1){
ll mid = (high + low) >> 1;
if(check2(mid)) low = mid;
else high = mid - 1;
//cout << "high = " << high << " low = " << low << endl;
}
if(check2(high)) return high;
return low;
}

int main()
{
IOS;
while(cin >> n >> k){
ll sum = 0;
for(int i = 0; i < n; i++){
cin >> data[i];
sum += data[i];
}
sort(data, data + n);
ll y = binsearch1();
ll x = binsearch2();
//cout << "y = " << y << " x = " << x << endl;
if(x >= y) {
if(sum % n == 0) cout << 0 << endl;
else cout << 1 << endl;
} else cout << y - x << endl;
}
return 0;
}