Codeforces Round #510 (Div. 2) D. Petya and Array

朱雀 2022-05-15 18:24 201阅读 0赞

## 题目：[点击打开链接][Link 1]   题意：给定一个数组，问有多少个不同的区间\[l,r\] (l<=r)使得区间和小于给定的数t。   分析：先求个前缀和，则问题转化为所有满足sum\[i\]-sum\[j\]<t（j<=i）的区间个数，原式可变形为-sum\[j\]<t-sum\[i\]，所以可以用一颗红黑树维护-sum\[j\]（[点击查看红黑树的库实现博客][Link 2]）,然后用order\_of\_key查询t-sum\[i\]的rank,这个rank其实就是右端点为i时左端点满足的个数。总复杂度nlog(n)，其中遍历数组为o(n),红黑树查询为log(n)。 ##

## 注：由于红黑树里不能有相同元素，所以这里让红黑树维护pair<-sum\[i\],index>,其中index为下标i.这样就能保证能存储相同的-sum\[i\].看红黑树实现起来多简短！ ##

## 补充：这题也可以用树状数组+map做，也是求前面前缀和大于sum\[i\]-t的个数，和树状数组求逆序数的思想一样，因为有负数，所以把前缀和数组统一加上一个大数变成正数，这个数要大于2e14，不过奇怪的是我把这个数设成奇数就超内存了，设成偶数就ac了，感觉用标准的离散化更保险。 ##

## 参考博客[https://blog.csdn.net/qq\_40791842/article/details/82780345][https_blog.csdn.net_qq_40791842_article_details_82780345]。   代码一（红黑树解法）： ##

#pragma comment(linker, "/STACK:102400000,102400000")
    #include<unordered_map>
    #include<unordered_set>
    #include<algorithm>
    #include<iostream>
    #include<fstream>
    #include<complex>
    #include<cstdlib>
    #include<cstring>
    #include<cassert>
    #include<iomanip>
    #include<string>
    #include<cstdio>
    #include<bitset>
    #include<vector>
    #include<cctype>
    #include<cmath>
    #include<ctime>
    #include<stack>
    #include<queue>
    #include<deque>
    #include<list>
    #include<set>
    #include<map>
    #include <ext/pb_ds/tree_policy.hpp>
    #include <ext/pb_ds/assoc_container.hpp>
    using namespace std;
    using namespace __gnu_pbds;
    #define pt(a) cout<<a<<endl
    #define debug test
    #define mst(ss,b) memset((ss),(b),sizeof(ss))
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,a,n) for (int i=n-1;i>=a;i--)
    #define pii pair<int,int>
    #define fi first
    #define se second
    #define ll long long
    #define ull unsigned long long
    #define pb push_back
    #define mp make_pair
    #define inf 0x3f3f3f3f
    #define eps 1e-10
    #define PI acos(-1.0)
    #define lb(x) (x&(-x))
    #define RBT tree<pair<ll,int>, null_type, less<pair<ll,int> >, rb_tree_tag, tree_order_statistics_node_update>
    const ll mod = 1e9+7;
    const int N = 1e6+10;
    
    ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    int to[4][2]={
       {-1,0},{1,0},{0,-1},{0,1}};
    
    ll n,t,x;
    
    int main() {
        ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
        while(~scanf("%lld%lld",&n,&t)) {
            RBT rbt;
            rbt.insert({0,0});
            ll ans=0,sum=0;
            for(int i=1;i<=n;i++) {
                scanf("%lld",&x);
                sum-=x;
                ans+=rbt.order_of_key({t+sum,0});
                rbt.insert({sum,i});
            }
            printf("%lld\n",ans);
        }
        return 0;
    }

## 代码二（树状数组解法）： ##

#pragma comment(linker, "/STACK:102400000,102400000")
    #include<unordered_map>
    #include<unordered_set>
    #include<algorithm>
    #include<iostream>
    #include<fstream>
    #include<complex>
    #include<cstdlib>
    #include<cstring>
    #include<cassert>
    #include<iomanip>
    #include<string>
    #include<cstdio>
    #include<bitset>
    #include<vector>
    #include<cctype>
    #include<cmath>
    #include<ctime>
    #include<stack>
    #include<queue>
    #include<deque>
    #include<list>
    #include<set>
    #include<map>
    using namespace std;
    #define pt(a) cout<<a<<endl
    #define debug test
    #define mst(ss,b) memset((ss),(b),sizeof(ss))
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,a,n) for (int i=n-1;i>=a;i--)
    #define pii pair<int,int>
    #define fi first
    #define se second
    #define ll long long
    #define ull unsigned long long
    #define pb push_back
    #define mp make_pair
    #define inf 0x3f3f3f3f
    #define eps 1e-10
    #define PI acos(-1.0)
    #define lb(x) (x&(-x))
    const ll mod = 1e9+7;
    const int N = 1e6+10;
    
    ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    int to[4][2]={
       {-1,0},{1,0},{0,-1},{0,1}};
    
    ll n,t,x,mx=2e14+10,a[N];
    map<ll,int> ma;
    
    void upd(ll x) {
        while(x<=2*mx) {
            ma[x]++;
            x+=lb(x);
        }
    }
    
    int gs(ll x) {
        int rs=0;
        while(x>0) {
            rs+=ma[x];
            x-=lb(x);
        }
        return rs;
    }
    
    int main() {
        ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
        cin>>n>>t;
        for(int i=1;i<=n;i++) cin>>a[i],a[i]+=a[i-1];
        upd(mx);
        ll ans=0;
        for(int i=1;i<=n;i++) {
            ans+=i-gs(a[i]+mx-t);
            upd(a[i]+mx);
        }
        cout<<ans<<endl;
        return 0;
    }

[Link 1]: http://codeforces.com/contest/1042/problem/D
[Link 2]: https://blog.csdn.net/riba2534/article/details/80454602
[https_blog.csdn.net_qq_40791842_article_details_82780345]: https://blog.csdn.net/qq_40791842/article/details/82780345