CodeForces 1151D Stas and the Queue at the Buffet

题目链接：http://codeforces.com/contest/1151/problem/D

题目大意：

　　有n个学生排成一队（序号从1到n），每个学生有2个评定标准(a, b)，设每个学生的位置为j，则每个学生所要交的学费为a * (j - 1) + b * (n - j)，要求把这些学生从新排序使得整体所交学费最小。

分析：

　　变换一下学费公式 = j * (a - b) + n * b - a，由于是求和，所以可以只看j * (a - b)部分，这就很显而易见了，(a - b)大的要排前面，(a - b)小的要排后面。

代码如下：

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
 
#define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "\n";
    return out;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef set< int > SI;
typedef vector< int > VI;
typedef map< int, int > MII;
const double EPS = 1e-10;
const int inf = 1e9 + 9;
const LL mod = 1e9 + 7;
const int maxN = 1e5 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

LL n, ans;

struct Student{
    LL a, b;
    
    LL disSa(int x) const {
        return a * (x - 1) + b * (n - x);
    }
};

Student stu[maxN];

bool cmp(const Student &x, const Student &y) {
    return x.a - x.b > y.a - y.b;
}

int main(){
    INIT();
    cin >> n; 
    For(i, 1, n) cin >> stu[i].a >> stu[i].b;
    
    sort(stu + 1, stu + n + 1, cmp);
    
    For(i, 1, n) ans += stu[i].disSa(i);
    cout << ans << endl;
    return 0;
}