Codeforces 672D Robin Hood

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题目链接：http://codeforces.com/problemset/problem/672/D

题意：对一个序列操作k次，每次将当前最大的数中分出去1给最小的数（如果有多个最大或最小随机选择，不过不影响答案）。求最后最大的数和最小的数的差值。

思路：k<=1e9 所以我们可以用二分法来做。现在我们假设操作k次后，最小的数为x。那么所有小于x的数都要加到x，t1 = ∑(x-a[i]) ( a[i] < x )，k至少要是这么多才可以；再考虑上界，如果操作很多，会把所有的x变成x+1或者更大，这样最小值就不是x了。所以我们记录所有小于等于x的数量t2，也就是说，当前最少操作t1次，这样所有小于x的数都会变成x，然后我们还可以最多再操作t2-1次，使得当前最小数还是x（再操作t2次，最小的数就会变成x+1）。二分求解最大的数同理。

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdlib>#include <iostream>#include <algorithm>#include <stack>#include <map>#include <set>#include <vector>#include <sstream>#include <queue>#include <utility>using namespace std;#define rep(i,j,k) for (int i=j;i<=k;i++)#define Rrep(i,j,k) for (int i=j;i>=k;i--)#define Clean(x,y) memset(x,y,sizeof(x))#define LL long long#define ULL unsigned long long#define inf 0x7fffffff#define mod %100000007const int maxn = 500009;LL n,k;LL a[maxn];void init(){    cin>>n>>k;    rep(i,1,n) scanf("%I64d",&a[i]);}int check( LL x , int type ){    LL t1 = 0;LL t2 = 0;rep(i,1,n)if ( type == 0 )    {        if ( a[i] <= x )        {            t1 += x - a[i];            t2++;        }    }    else    {        if ( a[i] >= x )        {            t1 += a[i] - x;            t2++;        }    }t2+=t1;if ( t1 > k ) return -1;if ( k >= t2 ) return 1;return 0;}void solve(){    LL S = 0;    LL m1,m2;    LL m,l,r;    LL ml,mr;    LL temp;ml = mr = a[1];    rep(i,1,n)    {        S += a[i];        if ( a[i] < ml ) ml = a[i];        if ( a[i] > mr ) mr = a[i];    }    m1 = m2 = -1;    l = ml;    r = S/n;    while( l < r )    {        m = ( l + r )>>1;        temp = check(m , 0);if (temp == -1) r = m - 1;else if ( temp == 1 ) l = m + 1;else{m1 = m;break;}    }    if (m1==-1) m1 = l;    l = (S+n-1)/n;    r = mr;    while( l < r )    {        m = ( l + r ) >> 1;        temp = check(m , 1);if (temp == -1) l = m + 1;else if ( temp == 1 ) r = m - 1;else{m2 = m;break;}    }    if (m2==-1) m2 = l;    printf("%I64d\n",m2-m1);}int main(){    init();    solve();    return 0;}

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