聪明的质检员

#include<cstdio>
#include<iostream>
#include<cstring>
#define LL long long
using namespace std;
LL sum[299999],cnt[299999];
int n,m;LL s;LL minn=99999999999;int ll[299999],rr[299999],w[299999],v[299999];
LL checkw(int x){
LL ans=0;
memset(sum,0,sizeof(sum));
memset(cnt,0,sizeof(cnt));
for(int i=1;i<=n;i++)
if(w[i]>=x) {
sum[i]=sum[i-1]+v[i];
cnt[i]=cnt[i-1]+1;
}
else cnt[i]=cnt[i-1],sum[i]=sum[i-1];//前缀和
for(int i=1;i<=m;i++)
{
ans+=(sum[rr[i]]-sum[ll[i]-1])*(cnt[rr[i]]-cnt[ll[i]-1]);//算y
}
minn=min(minn,((s-ans)>=0)?(s-ans):(ans-s));//找最优解
return ans;

}
int main(){

scanf("%d%d%lld",&n,&m,&s);
int l=1,r;
for(int i=1;i<=n;i++)
{
scanf("%d%d",&w[i],&v[i]);
r=max(r,w[i]);//找r的最大限度
}
r+=5;//防止二分出错
for(int i=1;i<=m;i++)
scanf("%d%d",&ll[i],&rr[i]);
while(l<=r)
{
int mid=(r+l)>>1;
LL y=checkw(mid);
if(y>s) l=mid+1;
else if(y<s) r=mid-1;
else {printf("0");return 0;}
}//二分答案

printf("%lld",minn);

return 0;
}