#1128 : 二分·二分查找 （ 两种方法 先排序在二分O(nlogN） + 直接二分+快排思想O(2N) ）

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9

//先排序在二分  48ms

#include<cstdio>
#include<algorithm>
//#include<bits/stdc++.h>
using namespace std;
template<class T>inline T read(T&x)
{
    char c;
    while((c=getchar())<=32)if(c==EOF)return 0;
    bool ok=false;
    if(c=='-')ok=true,c=getchar();
    for(x=0; c>32; c=getchar())
        x=x*10+c-'0';
    if(ok)x=-x;
    return 1;
}
template<class T> inline T read_(T&x,T&y)
{
    return read(x)&&read(y);
}
template<class T> inline T read__(T&x,T&y,T&z)
{
    return read(x)&&read(y)&&read(z);
}
template<class T> inline void write(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
template<class T>inline void writeln(T x)
{
    write(x);
    putchar('\n');
}
//-------ZCC IO template------
const int maxn=1000011;
const double inf=999999999;
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define For(i,t,n) for(int i=(t);i<(n);i++)
typedef long long  LL;
typedef double DB;
typedef pair<int,int> P;
#define bug printf("---\n");
#define mod 10007

LL a[maxn];
int bs(int left,int right,int key)
{
    while(left<=right)
    {
        int mid=(left+right)>>1;
        if(key==a[mid])return mid;
        else if(a[mid]>key)
            right=mid-1;
        else
            left=mid+1;
    }
    return -1;
}

int main()
{
    //#ifndef ONLINE_JUDGE
   // freopen("in.txt","r",stdin);
    //#endif // ONLINE_JUDGE
    int n,m,i,j,k,t;
    while(read_(n,k))
    {
        For(i,1,n+1)
        {
            read(a[i]);
        }
        sort(a+1,a+n+1);
        writeln(bs(1,n+1,k));
    }
    return 0;
}

//用二分的思想， 28ms

#include<cstdio>
#include<algorithm>
//#include<bits/stdc++.h>
using namespace std;
template<class T>inline T read(T&x)
{
    char c;
    while((c=getchar())<=32)if(c==EOF)return 0;
    bool ok=false;
    if(c=='-')ok=true,c=getchar();
    for(x=0; c>32; c=getchar())
        x=x*10+c-'0';
    if(ok)x=-x;
    return 1;
}
template<class T> inline T read_(T&x,T&y)
{
    return read(x)&&read(y);
}
template<class T> inline T read__(T&x,T&y,T&z)
{
    return read(x)&&read(y)&&read(z);
}
template<class T> inline void write(T x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
template<class T>inline void writeln(T x)
{
    write(x);
    putchar('\n');
}
//-------ZCC IO template------
const int maxn=1000011;
const double inf=999999999;
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define For(i,t,n) for(int i=(t);i<(n);i++)
typedef long long  LL;
typedef double DB;
typedef pair<int,int> P;
#define bug printf("---\n");
#define mod 10007
LL a[maxn];
int bs(int left,int right,int key)
{
    if(left>right)
    {
        if(a[left]==key)return left-1;
        else return -1;
    }
    int mid=a[left];
    int low=left;
    int high=right;
    while(low<high)
    {
        while(low<high&&a[high]>=mid)
            high--;
        if(low<high)swap(a[high],a[low]);
        while(low<high&&a[low]<=mid)
            low++;
        if(low<high)swap(a[high],a[low]);
    }
    if(mid<key)
        return bs(low+1,right,key);
    else
        return bs(left,low-1,key);
}



int main()
{
    //#ifndef ONLINE_JUDGE
   // freopen("in.txt","r",stdin);
    //#endif // ONLINE_JUDGE
    int n,m,i,j,k,t;
    while(read_(n,k))
    {
        For(i,1,n+1)
        {
            read(a[i]);
        }
        writeln(bs(1,n+1,k));
    }
    return 0;
}

#1128 : 二分·二分查找 （ 两种方法 先排序在二分O(nlogN） + 直接二分+快排思想O(2N) ）

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原文地址：http://blog.csdn.net/u013167299/article/details/45014951