HDU6383 p1m2

二分答案

可以发现一定是有解的，对于一个数组，每一次操作实际上是把和-1，最后一定可以变成一个01的序列

枚举最小值，然后统计上升和下降的次数。因为是统计最大的最小值，所以在二分答案偏小时统计答案。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int ret = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();
    return w ? -ret : ret;
}
inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template <typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template <typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 300005;
int _, n, a[N];

bool calc(int val){
    ll x = 0, y = 0;
    for(int i = 1; i <= n; i ++){
        if(a[i] > val) y += (a[i] - val) / 2;
        else if(a[i] < val) x += val - a[i];
    }
    return x <= y;
}

int main(){

    for(_ = read(); _; _ --){
        n = read();
        for(int i = 1; i <= n; i ++) a[i] = read();
        sort(a + 1, a + n + 1);
        int l = 0, r = 1e8;
        while(l < r){
            int mid = (l + r + 1) >> 1;
            if(calc(mid)) l = mid;
            else r = mid - 1;
        }
        printf("%d\n", l);
    }
    return 0;
}