数字游戏II

题面好难找：嘟嘟嘟

贪心 + dp。

首先要按bi的降序排序，让每一次减少大的数尽量靠前。为啥咧？于是我们就需要证明：令sum = a₁ - (1 - 1) * b₁ + a₂ - (2 - 1) * b₂ + a₃ - (3 - 1) * b₃ + ……+ a_n - (n - 1) * b_n，整理得：sum = (a₁ + a₂ + a₃ + ……+ a_n) + (0 * b₁ + 1 * b₂ + 2 * b₃ + ……+ (n - 1) * b_n)。因为Σa_n是定值，所以让sum最大就是让bi降序排序。

然后dp。

令dp[i][j]第 j 回合取ai的最大得分，仿照01背包：dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - 1] + a[i] - (j - 1) * b[i])。

于是这道题就做完啦。

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<vector>
 #include<stack>
 #include<queue>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a, x) memset(a, x, sizeof(a))
 #define rg register
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const db eps = 1e-;
 const int maxn = ;
 inline ll read()
 {
   ll ans = ;
   char ch = getchar(), last = ' ';
   while(!isdigit(ch)) {last = ch; ch = getchar();}
   while(isdigit(ch)) {ans = ans *  + ch - ''; ch = getchar();}
   if(last == '-') ans = -ans;
   return ans;
 }
 inline void write(ll x)
 {
   if(x < ) x = -x, putchar('-');
   if(x >= ) write(x / );
   putchar(x %  + '');
 }

 int n, m;
 struct Node
 {
   int a, b;
   bool operator < (const Node &oth)const
   {
     return b > oth.b;
   }
 }t[maxn];
 ll dp[maxn][maxn];

 int main()
 {
   n = read(); m = read();
   for(int i = ; i <= n; ++i) t[i].a = read();
   for(int i = ; i <= n; ++i) t[i].b = read();
   sort(t + , t + n + );
   for(int i = ; i <= n; ++i)
     for(int j = ; j <= m; ++j)
       dp[i][j] = max(dp[i - ][j], dp[i - ][j - ] + t[i].a - (j - ) * t[i].b);
   write(dp[n][m]), enter;
   return ;
 }