二分+前缀和

求解最小值，观察题目给出的判断条件，W随着变大时，区间中符合要求的会减少，同时所有，所以也随着变小，求得Y也是递减，可以得知W和Y是满足二分的关系。
在求解每一个时，运用到前缀和的思想O(1)实现。

总的时间复杂度

Code

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(vv) vv.begin(), vv.end()
#define endl "\n"
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 2e5 + 7;
int n, m;
ll s, w[N], v[N], l[N], r[N];
int cnt[N];
ll sum[N];

ll check(int x) {
    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    sum[i] = sum[i - 1], cnt[i] = cnt[i - 1];
    ll ans = 0;
    for (int i = 0; i < m; ++i)
        ans += (cnt[r[i]] - cnt[l[i] - 1]) * (sum[r[i]] - sum[l[i] - 1]);
    return ans;
}

int main() {
    n = read(), m = read(), s = read();
    for (int i = 1; i <= n; ++i)    w[i] = read(), v[i] = read();
    for (int i = 0; i < m; ++i)    l[i] = read(), r[i] = read();
    int l = 0, r = 1e6 + 7;
    while (l < r) {
        int mid = r + l + 1 >> 1;
        if (check(mid) >= s)    l = mid;
        else r = mid - 1;
    }
    write(min(abs(check(r) - s), abs(s - check(r + 1))));
    return 0;
}