P1314 聪明的质监员（前缀和+二分）

P1314 聪明的质监员

显然可以二分参数W

统计Y用下前缀和即可。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cctype>
 #include<algorithm>
 #include<cmath>
 #define re register
 using namespace std;
 typedef long long ll;
 void read(int &x){
     char c=getchar();x=;
     while(!isdigit(c)) c=getchar();
     while(isdigit(c)) x=(x<<)+(x<<)+(c^),c=getchar();
 }
 ll min(ll a,ll b){return a<b?a:b;}
 #define N 200002
 int n,m,w[N],v[N],tl[N],tr[N],c[N];
 ll s,sum1[N],sum2[N],ans=1e17;
 ll check(int lim){
     for(re int i=;i<=n;++i){
         if(w[i]<lim){
             sum1[i]=sum1[i-];
             sum2[i]=sum2[i-];
         }else{
             sum1[i]=sum1[i-]+v[i];//存权值和
             sum2[i]=sum2[i-]+;//存个数
         }
     }ll res=;
     for(re int i=;i<=m;++i){
         ll r1=sum1[tr[i]]-sum1[tl[i]-];
         ll r2=sum2[tr[i]]-sum2[tl[i]-];
         res+=r1*r2;//累加Y值
     }ans=min(ans,abs(res-s));
     return res;
 }
 int main(){
     read(n);read(m);scanf("%lld",&s);
     for(re int i=;i<=n;++i) read(w[i]),read(v[i]),c[i]=w[i];
     for(re int i=;i<=m;++i) read(tl[i]),read(tr[i]);
     sort(c+,c+n+);//重量值从小到大排序，用于二分
     int l=,r=n;
     while(l<r){
         int mid=l+((r-l)>>);
         if(check(c[mid])>s) l=mid+;//二分离s更近
         else r=mid;
     }printf("%lld",ans);
     return ;
 }