Codeforces 671B Robin Hood (二分搜索)

题意：

给定一个数列，每次让最大值-1，最小值+1,问你k次操作之后，数列中的最大值和最小值的差值是多少。

解法：

不管结果是怎样的，我们可以确定的是，最后的最大值和最小值肯定是固定的，但是我们不能确定到底是多少，所以我们需要尝试，而且随着操作次数的增加，最大值会越来越小，最小值会越来越大，所以我们采用二分搜索来找最后的值。值得注意的是，寻找最大值和最小值的边界是不一样的，具体在代码中。

代码：

//  Created by  CQU_CST_WuErli
//  Copyright (c) 2016 CQU_CST_WuErli. All rights reserved.
//
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <set>
#include <algorithm>
#include <sstream>
#define CLR(x) memset(x,0,sizeof(x))
#define OFF(x) memset(x,-1,sizeof(x))
#define MEM(x,a) memset((x),(a),sizeof(x))
#define BUG cout << "I am here" << endl
#define lookln(x) cout << #x << "=" << x << endl
#define SI(a) scanf("%d", &a)
#define SII(a,b) scanf("%d%d", &a, &b)
#define SIII(a,b,c) scanf("%d%d%d", &a, &b, &c)
const int INF_INT=0x3f3f3f3f;
const long long INF_LL=0x7f7f7f7f;
const int MOD=1e9+7;
const double eps=1e-10;
const double pi=acos(-1);
typedef long long  ll;
using namespace std;

ll n, k;
ll c[500050];

int main(int argc, char const *argv[]) {
#ifdef LOCAL
freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);
// freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);
#endif
while (scanf("%I64d%I64d", &n, &k) == 2) {
ll sum = 0;
for (int i = 1; i <= n; i++) {
scanf("%I64d", c + i);
sum += c[i];
}
ll L = 0, R = sum / n;
ll ans1 = 0;
while (L <= R) {
ll mid = L + R >> 1;
ll tmp = 0;
for (int i = 1; i <= n; i++)
if (c[i] < mid) tmp += mid - c[i];
if (tmp <= k) ans1 = mid, L = mid + 1;
else R = mid - 1;
}

L = (sum + n - 1) / n, R = 1000000000;
ll ans2 = 0;
while (L <= R) {
ll mid = L + R >> 1;
ll tmp = 0;
for (int i = 1; i <= n; i++)
if (c[i] > mid) tmp += c[i] - mid;
if (tmp <= k) ans2 = mid, R = mid - 1;
else L = mid + 1;
}
cout << ans2 - ans1 << endl;
}
return 0;
}