[CQOI2010]扑克牌

没想到省选题居然考的是二分答案！
首先，我们可以发现， 和 其实没什么区别，假如我们把 看成 0 号牌，那么，相当于这 种牌中任意 种各一张可以组成一套牌。

然后， 种各一张可以转化为 种各拿一张再取回去一张。

#include <algorithm>
#include <cctype>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstring>
#include <deque>
#include <functional>
#include <list>
#include <map>
#include <iomanip>
#include <iostream>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
#define int long long
#define I inline
#define ri register int
#define lowbit(x) x & -x
#define For(i , x , y) for(ri i = x ; i <= y ; ++ i)
#define Next(i , x) for(ri i = head[x] ; i ; i = e[i].nxt)
I int read() {
    int s = 0 , w = 1; char ch = getchar();
    while(ch < 48 || ch > 57) {if(ch == '-') w = -1; ch = getchar();}
    while(ch >= 48 && ch <= 57) s = (s << 1) + (s << 3) + (ch ^ 48) , ch = getchar();
    return s * w;
}

const int N = 105;
int n , m , a[N];

I bool judge(int mid) {
    int res = 0;
    For(i , 0 , n) res += max(0ll , mid - a[i]);
    return res <= mid;
}

signed main() {
    n = read() , a[0] = read();
    For(i , 1 , n) a[i] = read();
    int l = 0 , r = 1e10 , ans = 0; 
    while(l <= r) {
        int mid = (l + r) >> 1;
        if(judge(mid)) ans = mid , l = mid + 1;
        else r = mid - 1;
    } 
    cout << ans << endl;
    return 0;
}

``