Codeforces 903D

题意：

求：每个数和他前面的数，跟他差值大于一的数的差值的和

思路：

对于每一个数ai，减去他之前的所有的数和，这样的话 有一部分会多减，就是跟这个数差值是一的数，

这样我们再记录每一个数的数量，可以知道ai+1 和 ai-1 的数目，再相应的加上减去个数就是了

#include<iostream>#include<algorithm>#include<iomanip>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<cmath>#include<set>#include<queue>#include<stack>#include<map>#define PI acos(-1.0)#define in freopen("in.txt", "r", stdin)#define out freopen("out.txt", "w", stdout)#define kuaidian ios::sync_with_stdio(0);using namespace std;typedef long long ll;typedef unsigned long long ull;const int maxn = 2e5 + 7, maxd = 1e8;const ll mod = 1e9 + 7;const int INF = 0x7f7f7f7f;int n;ll sum[maxn];map<ll , int> mp;int main() {    kuaidian;    mp.clear();    cin >> n;        long double ans = 0;    for(int i = 1; i <= n; ++i) {        ll x;        cin >> x;        ans += ( (long double)(x*(i-1)) - (long double)(sum[i-1]) );        if(mp[x-1]) ans -= (long double)(mp[x-1]);        if(mp[x+1]) ans += (long double)(mp[x+1]);        sum[i] = sum[i-1] + x;        mp[x]++;    }    cout << fixed << setprecision(0) << ans << endl;        return 0;}