ABC:Meaningful Mean
题目描述
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
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
输入
Input is given from Standard Input in the following format:
N K
a1
a2
:
aN
N K
a1
a2
:
aN
输出
Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.
样例输入
3 6
7
5
7
样例输出
5
提示
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.
首先对于所有数减去k,这样就不用除(r-l+1), 然后我们发现所求的就是有多少对l,r,使得sum[r]-sum[l-1] >= 0, sum是减去k之后的序列的前缀和
用树状数组对sum求有多少个顺序对,多加一个0这个数,代表从sum[r]-sum[0]对答案的贡献。
由于sum[i]可能很大,所以需要离散化 用unique
#include <bits/stdc++.h> #define ll long long const int mod=1e9+7; const int maxn=2e5+7; using namespace std; //int s[maxn]; ll sum[maxn],t[maxn]; ll tree[maxn]; int nn,m; void add(int x) { while(x<=nn)//nn为上限,要大于离散化后的最大的数字, { tree[x]++; x+=x&-x;//向上更新,树状数组核心代码。此处就不解释了,不会的先学树状数组。 } } ll query(int x) { ll num=0; while(x) { num+=tree[x]; x-=x&-x;//向下求和 } return num; } int main() { int n,k,a; scanf("%d %d",&n,&k); nn=n+10; for(int i=1;i<=n;i++){ scanf("%d",&a); sum[i]=t[i]=sum[i-1]+a-k; } sort(t,t+n+1); m=unique(t,t+n+1)-t; for(int i=0;i<=n;i++){ sum[i]=lower_bound(t,t+m+1,sum[i])-t+1; } ll ans=0; for(int i=0;i<=n;i++){ ans+=query(sum[i]); add(sum[i]); } printf("%lld\n",ans); return 0; }
不要忘记努力,不要辜负自己
欢迎指正 QQ:1468580561