【CF1023B】Pair of Toys（解方程）

题意：给定n个玩具要你选出两个玩具求出k的价值，第i个玩具的价值为i。若是没有选择方案，输出0

补充：玩具A与玩具B 和 玩具B和玩具A 是同一种选择

n,k<=1e14

思路：列出式子，解不等式组

 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<queue>
 #include<vector>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef vector<int> VI;
 #define fi first
 #define se second
 #define MP make_pair
 #define N   210000
 #define MOD 1000000007
 #define eps 1e-8
 #define pi acos(-1)
 #define oo 2000000000

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 void swap(int &x,int &y)
 {
     int t=x;x=y;y=t;
 }

 ll max(ll x,ll y)
 {
     if(x>y) return x;
     return y;
 }

 ll min(ll x,ll y)
 {
     if(x<y) return x;
     return y;
 }

 int main()
 {
 //    freopen("1.in","r",stdin);
 //    freopen("1.out","w",stdout);
     ll n,k;
     scanf("%lld%lld",&n,&k);
      ll s=max(,max(k-n,k/+));
      ll t=min(n,k-);
      if(s>t) printf("");
       else
       {
           ll ans=t-s+;
         printf("%lld",ans);
       }

     return ;
 }