Codeforces976E Well played! 【Greedy】

Topic Analysis:

　　Since the gain of multiplying by two is very large, we can prove that the number of the multiplication must be the same, and then select it in turn after sorting, and judge it.

Topic code:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int maxn = 320000;
 5 
 6 int n,a,b;
 7 
 8 struct node{
 9     int hp,dm;
10 }d[maxn];
11 
12 int cmp(node alpha,node beta){
13     return alpha.hp-alpha.dm > beta.hp-beta.dm;
14 }
15 
16 void read(){
17     scanf("%d%d%d",&n,&a,&b);
18     for(int i=1;i<=n;i++) scanf("%d%d",&d[i].hp,&d[i].dm);
19     sort(d+1,d+n+1,cmp);
20 }
21 
22 void work(){
23     int hh = 1;
24     long long ext = 0,ans = 0;
25     for(int i=1;i<=a;i++) hh *= 2;
26     int pt;
27     for(pt=1;pt<=min(b,n);pt++){
28     if(d[pt].hp-d[pt].dm <= 0) break;
29     ans += d[pt].hp;
30     }
31     for(int i=pt;i<=n;i++) ans += d[i].dm;
32     long long em = ans;ext = ans;pt--;
33     if(b == 0){printf("%lld",ext);return;}
34     for(int i=1;i<=n;i++){
35     em = ans;
36     if(i<=pt){
37         em = em-d[i].hp+(1ll*d[i].hp*hh);
38         ext = max(ext,em);
39     }else{
40         if(pt < b){
41         em = em+(1ll*d[i].hp*hh)-d[i].dm;
42         ext = max(ext,em);
43         }else{
44         em = em-d[pt].hp+(1ll*d[i].hp*hh)-d[i].dm+d[pt].dm;
45         ext = max(ext,em);
46         }
47     }
48     }
49     printf("%lld",ext);
50 }
51 
52 int main(){
53     read();
54     work();
55     return 0;
56 }