Codeforces Round #510 (Div. 2) D. Petya and Array-蒲公英云

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

朱雀 2022-05-15 18:24 299阅读 0赞

题目：点击打开链接题意：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）,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。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;
}