AtCoder Beginner Contest 149 E.Handshake

AtCoder Beginner Contest 149 E.Handshake

Problem Statement

(略)
题目来源

Sample Input 1

5 3
10 14 19 34 33

Sample Output 1

202

Sample Output 2

1837

Sample Input 3

9 73
67597 52981 5828 66249 75177 64141 40773 79105 16076

Sample Output 3

8128170

比赛时根本想不到用二分，太菜了/(ㄒoㄒ)/~~，因为任意两数之和最大2e5，所以可以二分找到符合条件的最小和，最后再算一遍输出即可：

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int N=1e5+5;
ll sum[N],a[N],m,n;

int main()
{
    cin>>n>>m;
    for(ll i=1;i<=n;i++){
        cin>>a[i];
    }
    sort(a+1,a+n+1);
    for(ll i=1;i<=n;i++){
        sum[i]=sum[i-1]+a[i];
    }
    ll l=0,r=2e5+1,ans,res,cnt;
    while(l<=r){
        ll mid=(l+r)>>1;
        res=cnt=0;
        for(ll i=1;i<=n;i++){
            ll pos=lower_bound(a+1,a+n+1,mid-a[i])-a;
            cnt+=n-pos+1;
            res+=sum[n]-sum[pos-1]+a[i]*(n-pos+1);
        }
        if(cnt>=m){
            ans=mid;
            l=mid+1;
        }
        else r=mid-1;
    }
    printf("%lld\n",res-(cnt-m)*ans);
}