AtCoder 2581 Meaningful Mean（树状数组+离散化）

E - Meaningful Mean

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Time limit : 2sec / Memory limit : 256MB

Score : 600 points

Problem Statement

You are given an integer sequence of length N, a= {a1,a2,…,aN}, and an integer K.

a has N(N+1)⁄2 non-empty contiguous subsequences, {al,al+1,…,ar} (1≤l≤r≤N). Among them, how many have an arithmetic mean that is greater than or equal to K?

Constraints

• All input values are integers.
• 1≤N≤2×105
• 1≤K≤109
• 1≤ai≤109

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Input

Input is given from Standard Input in the following format:

N Ka1a2:aN

Output

Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.

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Sample Input 1

Copy

3 6757

Sample Output 1

Copy

5

All the non-empty contiguous subsequences of a are listed below:

• {a1} = {7}
• {a1,a2} = {7,5}
• {a1,a2,a3} = {7,5,7}
• {a2} = {5}
• {a2,a3} = {5,7}
• {a3} = {7}

Their means are 7, 6, 19⁄3, 5, 6 and 7, respectively, and five among them are 6 or greater. Note that {a1} and {a3} are indistinguishable by the values of their elements, but we count them individually.

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Sample Input 2

Copy

1 21

Sample Output 2

Copy

0

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Sample Input 3

Copy

7 2610203040302010

Sample Output 3

Copy

13

【思路】

这道题给了一串整数，要求的是算数平均数大于等于K的区间的数目。也就是a1，a2，a3，...，an，求任意[l，r]的区间和s（其中l <= r），若s/(r-l+1) >= K，则给答案加1。为了便于观察，我们将下标改为从0开始，用下面这种形式变化，可以得到br >= bl这种形式的式子，那么问题就转化成了求一对l和r（其中0 <= l < r <= n），使得bl <= br的l和r的对数，也就是求顺序对的对数。

很自然地就能想到用树状数组来维护，由于b可能非常大，所以要进行离散化。离散化的方式在这里有两种方式，

一是将b的值进行排序，从小到大扫一遍，找前面的数里头有多少下标小于等于当前数（由于下标不会重复，所以与严格小于是一样的），然后加起来，不过要尤其注意排序的时候若b值相等，要将下标小的排在前面，并且得注意处理0下标。

二是将b原序保留在s里，再对b排序去重离散化，然后把s扫一遍，找直到当前下标有多少个数小于等于当前数。

显然第二种方法又要排序，又要去重，又要保留原数组，所以时空复杂度都稍逊于第一种方法，但是第一种方法较易出错。我自己先写的第一种总是WA，看了同学代码后写出了第二种，然后突然明白第一种应该要双关键字排序，又改了改才对了。这里也附上两种写法。

【代码1】

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN=200005;struct node{    long long num;    int id;    bool operator<(const node &another)const    {        if(num==another.num)return id<another.id;        return num<another.num;    }};int n,k;long long a[MAXN];node b[MAXN];int lowbit(int x){    return (x&-x);}void modify(int x,int num){    if(x==0){        a[x]+=(long long)num;        return;    }    while(x<=n){        a[x]+=(long long)num;        x+=lowbit(x);    }}long long sum(int x){    long long ans=a[0];    while(x>0){        ans+=a[x];        x-=lowbit(x);    }    return ans;}int main(){    scanf("%d %d",&n,&k);    b[0].num=0;b[0].id=0;    for(int i=1;i<=n;i++){        int x;        scanf("%d",&x);        b[i].num=b[i-1].num+(long long)(x-k);        b[i].id=i;    }    sort(b,b+1+n);    long long cnt=0;    memset(a,0,sizeof(a));    for(int i=0;i<=n;i++){        if(b[i].id>0)cnt+=sum(b[i].id-1);        modify(b[i].id,1);    }    printf("%lld\n",cnt);    return 0;}

【代码2】

#include<cstdio>#include<cstring>#include<algorithm>#include<map>using namespace std;const int MAXN=200005;int n,k;long long a[MAXN],s[MAXN],b[MAXN];map<long long,int> mp;int lowbit(int x){    return (x&-x);}void modify(int x,int num){    while(x<=n+1){        a[x]+=(long long)num;        x+=lowbit(x);    }}long long sum(int x){    long long ans=0;    while(x>0){        ans+=a[x];        x-=lowbit(x);    }    return ans;}int main(){    scanf("%d %d",&n,&k);    s[0]=0;b[0]=0;    for(int i=1;i<=n;i++){        int x;        scanf("%d",&x);        s[i]=s[i-1]+(long long)(x-k);        b[i]=s[i];    }    sort(b,b+1+n);    int tot=0;    for(int i=0;i<=n;i++)        if(!mp.count(b[i]))mp[b[i]]=++tot;    long long cnt=0;    memset(a,0,sizeof(a));    for(int i=0;i<=n;i++){        cnt+=sum(mp[s[i]]);        modify(mp[s[i]],1);    }    printf("%lld\n",cnt);    return 0;}