cf 760B.Frodo and pillows

二分，判断条件就是最小情况（设当前k位取x）比剩余值（m-x）要小。（貌似又做麻烦了2333）

 #include<bits/stdc++.h>
 #define LL long long
 #define  N 100005
 using namespace std;
 inline int ra()
 {
     int x=,f=; char ch=getchar();
     while (ch<'' || ch>'') {if (ch=='-') f=-; ch=getchar();}
     while (ch>='' && ch<='') {x=x*+ch-''; ch=getchar();}
     return x*f;
 }
 int n,m,k;
 bool check(LL x)
 {
     if (x==) return m>=n;
     LL mx1=(k-)*(x++x+k-)>>,mn1;
     LL mx2=(n-k)*(x++x+n-k)>>,mn2;
     if (x>=k) mn1=(k-)*(x+x-k)>>;
     else mn1=((x-)*(x)>>)+(k-x);
     if (x>=n-k) mn2=(n-k)*(x-+x-n+k)>>;
     else  mn2=((x-)*(x)>>)+(n-k-x+);
     if (mn1+mn2>m-x) return ;
     return ;
 }
 int main()
 {
     n=ra(); m=ra(); k=ra();
     int l=,r=m;
 //    check(3); while (1);
     while (l<r)
     {
         int mid=l+r>>;
         if (check(mid)) l=mid+;
         else r=mid;
     }
     if (check(l)) cout<<l; else cout<<l-;
     return ;
 }