思路：

　　二分；

 

代码：

#include 
     
      using namespace std;#define maxn 200005#define ll long long#define INF 0x7fffffffll n,m,sum[maxn],wi[maxn],vi[maxn],s;ll li[maxn],ri[maxn],sumn[maxn],flag;inline void in(ll &now){    char Cget=getchar();now=0;    while(Cget>'9'||Cget<'0')Cget=getchar();    while(Cget>='0'&&Cget<='9')    {        now=now*10+Cget-'0';        Cget=getchar();    }}int solve(ll val){    sum[0]=0,sumn[0]=0;    for(ll i=1;i<=n;i++)    {        sum[i]=sum[i-1],sumn[i]=sumn[i-1];        if(wi[i]>=val) sum[i]+=vi[i],sumn[i]++;    }    ll res=0;    for(ll i=1;i<=m;i++) res+=(sumn[ri[i]]-sumn[li[i]-1])*(sum[ri[i]]-sum[li[i]-1]);    if(res>=s)    {        flag=res-s;        return false;    }    else    {        flag=s-res;        return true;    }}int main(){    in(n),in(m),in(s);    for(ll i=1;i<=n;i++) in(wi[i]),in(vi[i]);    for(ll i=1;i<=m;i++) in(li[i]),in(ri[i]);    ll l=0,r=1000000,ans,mid;    while(l<=r)    {        mid=l+r>>1;        if(solve(mid)) ans=mid,r=mid-1;        else l=mid+1;    }    solve(ans);ll ansl=flag;    solve(ans-1);ll ansr=flag;    printf("%lld",min(ansl,ansr));    return 0;}