hiho第三十六周

二分查找
1.利用STL的lower_bound

#include <bits/stdc++.h>
#define maxn 1000010
using namespace std;
int a[maxn];
int main()
{
int n, k;

scanf( "%d%d", &n, &k);
for ( int i = 0; i < n; i++){
scanf( "%d", &a[i]);
}
sort( a, a + n);
int * x = lower_bound( a, a + n, k);
if ( *x == k)
printf( "%d\n", x - a + 1);
else
printf( "-1\n");
return 0;
}

2.直接写二分
#include <bits/stdc++.h>
#define maxn 1000010
using namespace std;
int a[maxn];
int main()
{
int n, k;

scanf( "%d%d", &n, &k);
for ( int i = 0; i < n; i++)
scanf( "%d", &a[i]);
sort( a, a + n);
int x = 0, y = n - 1, mid;
while ( x <= y){
mid = ( x + y) / 2;
if ( a[mid] == k)
break;
else if ( a[mid] > k)
y = mid - 1;
else if ( a[mid] < k)
x = mid + 1;
}
if ( x <= y)
printf( "%d\n", mid + 1);
else
printf( "-1\n");
return 0;
}