此题难点在于前缀和的定义 只计算符合条件（w[i] > W)部分的前缀和
二分 + 前缀和
理解题意 容易得出 当W递增 则y递减 因为符合要求的矿产数量会减少
二分参数 W
此处的前缀和是非连续前缀和 以及 计数前缀和

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 2e5 + 10;
int n, m;
ll s;
ll cnt[N], qzv[N];
ll sum;
int qjl[N], qjr[N];
ll ttt;

struct md{
    int w, v;
}kc[N];

//bool cmp(md x, md y)
//{
//  return x.w < y.w;
//}

bool check(int x)
{
//  int l = 1, r = n;
//  while(l < r)
//  {
//      int mid = (l + r) >> 1;
//      if(kc[mid].w >= x)r = mid;
//      else l = mid + 1;
//  }

    memset(cnt, 0, sizeof cnt);
    memset(qzv, 0, sizeof qzv);

    for(int i = 1;i <= n;i++)
    {
        if(kc[i].w >= x)cnt[i] = cnt[i - 1] + 1, qzv[i] = qzv[i - 1] + kc[i].v;
        else cnt[i] = cnt[i - 1], qzv[i] = qzv[i - 1];
//      cout << cnt[i] << " " << qzv[i] << endl;
    }

    ll y = 0;
    for(int i = 1;i <= m;i++)
    {
        y += (cnt[qjr[i]] - cnt[qjl[i] - 1]) * (qzv[qjr[i]] - qzv[qjl[i] - 1]); 
    }
    ttt = abs(s - y);
    if(y <= s)return true;
    return false;
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    cin >> n >> m >> s;
    for(int i = 1;i <= n;i++){
        cin >> kc[i].w >> kc[i].v;
    }

    for(int i = 1;i <= m;i++)cin >> qjl[i] >> qjr[i];

    int l = 1, r = 1000000;
    ll ans = 0x3f3f3f3f3f3f3f3f;
    while(l < r)
    {
        int mid = (l + r) >> 1;
        if(check(mid))r = mid;
        else l = mid + 1;
        ans = min(ans, ttt);
    }
    cout << ans << endl;
    return 0;
}