题目链接：

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：

1 #pragma GCC optimize("Ofast") 2 #include 
      
        3 using namespace std; 4   5 #define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0); 6 #define Rep(i,n) for (int i = 0; i < (n); ++i) 7 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 8 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 9 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)10 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)11 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)12 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)13  14 #define pr(x) cout << #x << " = " << x << "  "15 #define prln(x) cout << #x << " = " << x << endl16  17 #define LOWBIT(x) ((x)&(-x))18  19 #define ALL(x) x.begin(),x.end()20 #define INS(x) inserter(x,x.begin())21  22 #define ms0(a) memset(a,0,sizeof(a))23 #define msI(a) memset(a,inf,sizeof(a))24 #define msM(a) memset(a,-1,sizeof(a))25 26 #define MP make_pair27 #define PB push_back28 #define ft first29 #define sd second30  31 template
       
        32 istream &operator>>(istream &in, pair
        
          &p) {33     in >> p.first >> p.second;34     return in;35 }36  37 template
         
          38 istream &operator>>(istream &in, vector
          
            &v) {39     for (auto &x: v)40         in >> x;41     return in;42 }43  44 template
           
            45 ostream &operator<<(ostream &out, const std::pair
            
              &p) {46 out << "[" << p.first << ", " << p.second << "]" << "\n";47 return out;48 }49 50 typedef long long LL;51 typedef unsigned long long uLL;52 typedef pair< double, double > PDD;53 typedef pair< int, int > PII;54 typedef set< int > SI;55 typedef vector< int > VI;56 typedef map< int, int > MII;57 const double EPS = 1e-10;58 const int inf = 1e9 + 9;59 const LL mod = 1e9 + 7;60 const int maxN = 1e5 + 7;61 const LL ONE = 1;62 const LL evenBits = 0xaaaaaaaaaaaaaaaa;63 const LL oddBits = 0x5555555555555555;64 65 LL n, ans;66 67 struct Student{68 LL a, b;69 70 LL disSa(int x) const {71 return a * (x - 1) + b * (n - x);72 }73 };74 75 Student stu[maxN];76 77 bool cmp(const Student &x, const Student &y) {78 return x.a - x.b > y.a - y.b;79 }80 81 int main(){82 INIT();83 cin >> n; 84 For(i, 1, n) cin >> stu[i].a >> stu[i].b;85 86 sort(stu + 1, stu + n + 1, cmp);87 88 For(i, 1, n) ans += stu[i].disSa(i);89 cout << ans << endl;90 return 0;91 }