$Sol$
First, it must be a $ W $ $ w_i $. Then it came out a way to violence, and enumerate $ W $ recalculation.
He noted that, to meet the SY $ $ $ of the smallest absolute value of the Y $ only be two, one is $ <S $ maximum the Y $ $ A is $> S $ minimum $ Y $. That They are seeking out pair of. values were seeking when the $ W $ is to be dichotomous. but this does not $ a $ out this problem, because the $ check $ complexity is still $ O (NM) $ the look after explanations can be found $ $ prefix and Check acridine, I feel very clever $ qwq $. such complexity down $ Check () becomes $ $ O (N + M). $
$Code$
#include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<algorithm> #define il inline #define Rg register #define go(i,a,b) for(Rg int i=a;i<=b;++i) #define yes(i,a,b) for(Rg int i=a;i>=b;--i) #define mem(a,b) memset(a,b,sizeof(a)) #define int long long #define db double #define inf 2147483647 using namespace std; il int read() { Rg int x=0,y=1;char c=getchar(); while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();} while(c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-'0';c=getchar();} return x*y; } const int N=200010; int n,m,S,as,minw=inf,maxw,sn[N],sv[N]; struct nd1{int w,v;}a[N]; struct nd2{int l,r;}b[N]; il int calc(int x) { Rg int ret=0; mem(sn,0);mem(sv,0); go(i,1,n) if(a[i].w>=x)sn[i]=sn[i-1]+1,sv[i]=sv[i-1]+a[i].v; else sn[i]=sn[i-1],sv[i]=sv[i-1]; go(i,1,m) { Rg int l=b[i].l,r=b[i].r; ret+=(sn[r]-sn[l-1])*(sv[r]-sv[l-1]); } return ret; } il int ef1() { Rg int l=minw,r=maxw,mid,y,ret; while(l<=r) { mid=(l+r)>>1; y=calc(mid); if(y<=S)ret=y,r=mid-1; else l=mid+1; } return ret; } il int ef2() { Rg int l=minw,r=maxw,mid,y,ret; while(l<=r) { mid=(l+r)>>1; y=calc(mid); if(y>=S)ret=y,l=mid+1; else r=mid-1; } return ret; } main() { n=read(),m=read(),S=read(); go(i,1,n)a[i]=(nd1){read(),read()},minw=min(minw,a[i].w),maxw=max(maxw,a[i].w); go(i,1,m)b[i]=(nd2){read(),read()}; Rg int y1=ef1(),y2=ef2(); as=min(abs(y1-S),abs(y2-S)); printf("%lld\n",as); return 0; }