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-;
 const int inf = 1e9 + ;
 const LL mod = 1e9 + ;
 const int maxN = 1e5 + ;
 const LL ONE = ;
 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 const LL oddBits = 0x5555555555555555;

 LL n, ans;

 struct Student{
     LL a, b;

     LL disSa(int x) const {
         return a * (x - ) + 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, , n) cin >> stu[i].a >> stu[i].b;

     sort(stu + , stu + n + , cmp);

     For(i, , n) ans += stu[i].disSa(i);
     cout << ans << endl;
     return ;
 }