hiho#1128 : 二分·二分查找

input

1<=n<=1e6 1<=k<=2*1e9

a1 a2 ... an 1<=an<=2*1e9

output

k存在则输出k是第几大的数，否则输出-1

做法，从两端开始找到中间，把比b大的和比b小的交换

 #include <cstdio>
 #include <queue>
 #include <cstring>
 #include <iostream>
 #include <cstdlib>
 #include <algorithm>
 #include <vector>
 #include <map>
 #include <set>
 #include <ctime>
 #include <cmath>

 using namespace std;

 const int N=1e6;
 int a[N+],n,b;

 int main()
 {
     //freopen("/home/user/桌面/in","r",stdin);
     while(scanf("%d%d",&n,&b)==)
     {
         for(int i=;i<n;i++)
             scanf("%d",&a[i]);
         int i=,j=n-,find=;
         while()
         {
             for(;i<j;i++)
                 if(a[i]>=b) break;
             for(;j>i;j--)
                 if(a[j]<b) break;
                 else if(a[j]==b) find=;//else少了会WA
             if(a[i]==b) find=;
             if(i>=j) break;
             swap(a[i],a[j]);
         }
         find?printf("%d\n",i+):puts("-1");
     }
     //printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
     return ;
 }