提出 #4634379

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ソースコード 拡げる

#include<algorithm>
#include<complex>
#include<ctype.h>
#include<iomanip>
#include<iostream>
#include<fstream>
#include<map>
#include<math.h>
#include<numeric>
#include<queue>
#include<set>
#include<stack>
#include<stdio.h>
#include<string>
#include<string>
#include<vector>

using namespace std;
typedef long long ll;

#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define ALL(v) (v).begin(), (v).end()
#define p(s) cout<<(s)<<endl
#define p2(s, t) cout << (s) << " " << (t) << endl
#define br() p("")
#define pn(s) cout << (#s) << " " << (s) << endl
#define p_yes() p("Yes")
#define p_no() p("No")

const ll mod = 1e9 + 7;
const ll inf = 1e18;

template < typename T >
void vprint(T &V){
	for(auto v : V){
    	cout << v << " ";
	}
	cout << endl;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    // input
    ll N, D;
    cin >> N >> D;
    
    vector<ll> A(N);
    FOR(i, 0, N){
        cin >> A.at(i);
    }

    ll sum = 0;
    // 中央jから見る
    FOR(j, 1, N-1){
        auto it_j = lower_bound(ALL(A), A[j]);

        // 右
        auto it_k = lower_bound(ALL(A), A[j]+D);
        if(it_k==A.end()){
            it_k--;
        }
        else if(*it_k!=A[j]+D){
            it_k--;
        }
        ll right = it_k - it_j;

        // a_i
        auto it_i = lower_bound(ALL(A), A[j]-D);
        ll left = it_j - it_i;

        sum += left * right;
    }

    // 余分なのを除去
    FOR(i, 0, N-2){
        auto it_i = lower_bound(ALL(A), A[i]);
        auto it = lower_bound(ALL(A), A[i]+D);
        if(it==A.end()){
            it--;
        }
        else if(*it!=A[i]+D){
            it--;
        }
        ll num = it - it_i;
        if(num>=2){
            ll excess = num * (num-1) / 2;
            sum -= excess;
        }
    }

    p(sum);
    
    return 0;
}

提出情報

ジャッジ結果