hiho第三十六週

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.直接寫二分

#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;
}