玲珑OJ 1149

题目链接

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显然通过尺取法可以找到每次以j为结尾,包含j的区间,往左延伸,最左能到i (i<=j)
于是以j为结尾的区间有(j-i+1)个
累加即可

#include<stdio.h>#include<bits/stdc++.h>#define ll long long#define pii pair<int,int>#define pll pair<ll,ll>#define MEM(a,x) memset(a,x,sizeof(a))#define lowbit(x) ((x)&-(x))using namespace std;const int N = 2e5+5;int a[N];map<int,int>mp;void eraseAi(int i){    auto it=mp.find(a[i]);    --it->second;    if(it->second==0){        mp.erase(it);    }}ll slove(int n,int k){    mp.clear();    int i=0;    mp[a[i]]=1;    ll ans=1;    for(int j=1;j<n;++j){        mp[a[j]]+=1;        while((--mp.end())->first-mp.begin()->first>k){            eraseAi(i);            ++i;        }        ans+=(j-i+1);    }    return ans;}int main(){    //freopen("/home/lu/code/r.txt","r",stdin);    //freopen("/home/lu/code/w1.txt","w",stdout);    int n,k;    while(~scanf("%d%d",&n,&k)){        for(int i=0;i<n;++i){            scanf("%d",&a[i]);        }        printf("%lld\n",slove(n,k));    }    return 0;}