POJ 3579，poj 3685解题记录

二分中值，检查的时候继续二分，二分套二分。。。。。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<functional>
#include<queue>
using namespace std;
typedef long long ll;
const int maxv=1e5+30;
ll n;
ll tol;
ll a[maxv];
bool check(ll mid){
    ll num=0;
    for(int i=0;i<n;i++){ 
        if(a[i]+mid>a[n-1]) break;
        ll *p=lower_bound(a+i,a+n,a[i]+mid);
        num+=n-(p-a);
    }
    if(num>(tol/2)) return 1;
    else return 0;
}
int main(){
    freopen("in","r",stdin);
    while(cin>>n){
        tol=n*(n-1)/2;
        for(int i=0;i<n;i++) scanf("%lld",&a[i]);
        sort(a,a+n);
        ll l=0,r=1e9;
        while(r-l>1){
            ll mid=(r+l)/2;
            if(check(mid)) l=mid;
            else r=mid;
        }
        if(check(l)) cout<<l<<endl;
        else cout<<r<<endl;
    }
    return 0;
}

 poj 3685和这个差不多也贴在这里，感觉二分写得还是有问题。。。老是wa

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<functional>
#include<queue>
using namespace std;
typedef long long ll;
const int maxv=5e4+30;
ll T,N,m;
ll cul(const ll &i,const ll &j){
    return i*i+(ll)(1e5)*i+j*j-(ll)(1e5)*j+i*j;
}
bool check(ll &k){
    ll num=0;///小于k的数量
    for(int j=N;j>=1;j--){
        ll r=N+1,l=0;
        while(r-l>1){
            ll mid=(r+l)>>1;
            if(cul(mid,j)<k) l=mid;
            else r=mid;
        }
        num+=l;
    }
    return num<m;
}
int main(){
    //freopen("in","r",stdin);
    cin>>T;
    while(T--){
        cin>>N>>m;
        ll l= -100000 * N,r=N * N + 100000 * N + N * N + N * N;
        while(r-l>1){
            ll mid=(r+l)>>1;
            if(check(mid)) l=mid;
            else r=mid;
        }
        cout<<l<<endl;
    }
    return 0;
}