AcWing每日一题（提高组）--聪明的质监员

https://www.acwing.com/problem/content/501/

 1 #include<iostream>
 2 using namespace std;
 3 const int N=200010;
 4 typedef long long LL;
 5 int v[N],w[N];
 6 int L[N],R[N];
 7 int n,m;
 8 LL S;
 9 LL sum[N],cnt[N];
10 LL get_Y(int W){
11     for(int i=1;i<=n;i++){
12         if(w[i]>=W){
13             sum[i]=sum[i-1]+v[i];
14             cnt[i]=cnt[i-1]+1;
15         }else{
16             sum[i]=sum[i-1];
17             cnt[i]=cnt[i-1];
18         }
19     }
20     LL res=0;
21     for(int i=1;i<=m;i++){
22         res += (sum[R[i]] - sum[L[i] - 1]) * (cnt[R[i]] - cnt[L[i] - 1]);
23     }
24     return res;
25 }
26 int main(void){
27     cin>>n>>m>>S;
28     for(int i=1;i<=n;i++){
29         cin>>w[i]>>v[i];
30     }
31     for(int i=1;i<=m;i++){
32         cin>>L[i]>>R[i];
33     }
34     int l=0,r=1e6+1;
35     while(l<r){
36         int mid=l+r+1>>1;
37         if(get_Y(mid)>=S){
38             l=mid;
39         }else{
40             r=mid-1;
41         }
42     }
43     cout<<min(get_Y(r)-S,S-get_Y(r+1));
44     return 0;
45 }

标签：cnt,--,sum,质监,mid,long,int,LL,AcWing
来源： https://www.cnblogs.com/greenofyu/p/14397883.html