```cpp
#include<bits/stdc++.h>
using namespace std;
const int maxn=200010;
int w[maxn],v[maxn],l[maxn],r[maxn];
long long pre_n[maxn],pre_v[maxn];
long long Y,s,sum;
int n,m,mx=-1,mn=2147483647;
bool check(int W)
{
Y=0,sum=0;
memset(pre_n,0,sizeof(pre_n));
memset(pre_v,0,sizeof(pre_v));
for(int i=1;i<=n;i++)
{
if(w[i]>=W) pre_n[i]=pre_n[i-1]+1,pre_v[i]=pre_v[i-1]+v[i];
else pre_n[i]=pre_n[i-1],pre_v[i]=pre_v[i-1];
}
for(int i=1;i<=m;i++)
Y+=(pre_n[r[i]]-pre_n[l[i]-1])*(pre_v[r[i]]-pre_v[l[i]-1]);
sum=llabs(Y-s);
if(Y>s) return true;
else return false;
}
int main(){
// freopen("qc.in","r",stdin);
// freopen("qc.out","w",stdout);
scanf("%d %d %lld",&n,&m,&s);
for(int i=1;i<=n;i++)
{
scanf(" %d %d",&w[i],&v[i]);
mx=max(mx,w[i]);
mn=min(mn,w[i]);
}
for(int i=1;i<=m;i++)
scanf(" %d %d",&l[i],&r[i]);
int left=mn-1,right=mx+2,mid; //这里有的人说要特判左右端点的check，但是其实你把left开成mn-1,right开成mx+2(注意取mx+1时即为W比所有都大，Y是0，这个情况要考虑，所以+2包含mx+1)就可以包含左右端点的check了，会简单点。
long long ans=999999999999999;
while(left<=right)
{
mid=(left+right)>>1;
if(check(mid)) left=mid+1;
else right=mid-1;
if(sum<ans) ans=sum;
}
printf("%lld",ans);
return 0;
}
```
by ☆Ork_Morry☆✔ @ 2018-11-11 21:06:21