题目描述

点进去看吧，说的不能再清楚了。

Solution

看到数据规模不难想到二分 $W$W，然后用个前缀和优化一下即可。注意上下界。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

using namespace std;
#define int long long
const int MAXN=200010;

int n,m,s;
int maxx=-1,minn=0x3f3f3f3f;
int ans=(int)1e17l;
int Y,cnt;
int sumn[MAXN],sumv[MAXN];
int v[MAXN],w[MAXN];
int l[MAXN],r[MAXN];

int check(int x){
	Y=cnt=0;
	memset(sumn,0,sizeof(sumn));
	memset(sumv,0,sizeof(sumv));
	for(int i=1;i<=n;i++)
	{
		if(w[i]>=x) sumn[i]=sumn[i-1]+1,sumv[i]=sumv[i-1]+v[i];
		else sumn[i]=sumn[i-1],sumv[i]=sumv[i-1];
	}
	for(int i=1;i<=m;i++)
		Y+=(sumn[r[i]]-sumn[l[i]-1])*(sumv[r[i]]-sumv[l[i]-1]);
	cnt=abs(Y-s);
	return Y>s;
}
inline int read(){
	int x=0; char c;
	do c=getchar(); while(c<'0'||c>'9');
	while(c>='0'&&c<='9')
		x=x*10+c-48,c=getchar();
	return x;
}
signed main(){
	n=read();m=read();s=read();
	for(int i=1;i<=n;++i){
		w[i]=read();v[i]=read();
		maxx=max(maxx,w[i]);minn=min(minn,w[i]);
	}
	for(int i=1;i<=m;++i){
		l[i]=read();
		r[i]=read();
	}
	int left=minn-1,right=maxx+2,mid;
	while(left<=right){
		mid=(left+right)/2;
		if(check(mid))
			left=mid+1;
		else
			right=mid-1;
		ans=min(ans,cnt);
	}
	printf("%lld",ans);
}