Codeforces 903D Almost Difference

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题意：

定义一个函数

给定一个含有n个数的数组a，要求求出所有d(ai,aj)的和，其中1<=i<=j<=n。

其中1<=n<=2e5，1<=ai<=1e9。

题解：

虽然ai最大为1e9，但最大也就只有2e5个数，因此可以先离散化再建线段树。线段树里每个节点存有数的和还有数的个数。遍历每个数，对于当前的这个数ai，求出比它小的数的和sum和个数count，则count*ai-sum加到结果里，其中注意ai在线段树的前一个叶子节点是不是ai-1，是的话要把这个点忽略掉。接着求出比它大的数的和sum和个数count，接下来同理。最后累加得到的就是结果。结果会爆longlong，可以用long double，但是注意一下输出格式。

#include<bits/stdc++.h>using namespace std;#define ls t<<1#define rs (t<<1)|1const int MAXN=200010;const int MAXN1=2001000;const int INF=0x3f3f3f3f;typedef long long ll;typedef long double ld;int n,cnt;ll a[MAXN],b[MAXN];map<ll,int> ma;int fa[MAXN];struct NODE{    int left, right,lazy,sum;    ld val;}tree[MAXN1];void build(int left, int right, int t){tree[t].left = left;tree[t].right = right;tree[t].val = tree[t].lazy = tree[t].sum=0;if (left == right){fa[left] = t;return;}build(left, (left + right) / 2, ls);build((left + right) / 2 + 1, right, rs);}void update1(int t){if (t == 1) return;t = t / 2;    tree[t].val = tree[ls].val + tree[rs].val;    tree[t].sum=tree[ls].sum+tree[rs].sum;update1(t);}ld queryVal(int left, int right, int t){ld sum = 0;    if (tree[t].left >= left&&tree[t].right <= right)return tree[t].val;int mid = (tree[t].left + tree[t].right) / 2;if (left <= mid) sum += queryVal(left, right, ls);if (right > mid) sum += queryVal(left, right, rs);return sum;}int querySum(int left, int right, int t){int sum = 0;    if (tree[t].left >= left&&tree[t].right <= right)return tree[t].sum;int mid = (tree[t].left + tree[t].right) / 2;if (left <= mid) sum += querySum(left, right, ls);if (right > mid) sum += querySum(left, right, rs);return sum;}int main(){ //   freopen("input.txt","r",stdin); //   freopen("output.txt","w",stdout);    ios_base::sync_with_stdio(0); cin.tie(0);    cin>>n;    for(int i=1;i<=n;i++)    {        cin>>a[i];        b[i]=a[i];    }    sort(b+1,b+1+n);    for(int i=1;i<=n;i++)    {        if(ma.find(b[i])==ma.end())         {            cnt++;            ma[b[i]]=cnt;        }    }    build(1,cnt,1);    ld ans=0;    for(int i=1;i<=n;i++)    {        int ord=ma[a[i]];        tree[fa[ord]].val+=a[i];        tree[fa[ord]].sum++;        update1(fa[ord]);        ll val;        int sum;        if(ord!=1)         {            if(ma.find(a[i]-1)==ma.end())             {                val=queryVal(1,ord-1,1);                sum=querySum(1,ord-1,1);                ans+=sum*a[i]-val;            }            else             {                if(ord!=2)                {                    val=queryVal(1,ord-2,1);                    sum=querySum(1,ord-2,1);                    ans+=sum*a[i]-val;                }            }        }        if(ord!=cnt)        {            if(ma.find(a[i]+1)==ma.end())             {                val=queryVal(ord+1,cnt,1);                sum=querySum(ord+1,cnt,1);                ans+=sum*a[i]-val;            }            else if(ord!=cnt-1)            {                val=queryVal(ord+2,cnt,1);                sum=querySum(ord+2,cnt,1);                ans+=sum*a[i]-val;            }        }    }    cout << fixed << setprecision(0) << ans << endl;    return 0;}

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