仔细读懂数据生成器，就能看出来，实际上物品肯定是够用的。

因为只能从右向左搬运物品，所以我们只需要对于每一个$i,i+1$的间隔，考虑有多少个物资需要从右边搬到左边去，把这个贡献累加即可，时间复杂度$O(n)$。

#include <bits/stdc++.h>
#define maxn 10000005
#define ll long long
using namespace std;
ll n;
ll c[maxn];
int a[maxn], b[maxn];
int randseed;
unsigned int rnd() {
    unsigned int r;
    r = randseed = randseed * 1103515245 + 12345;
    return (r << 16) | ((r >> 16) & 0xFFFF);
}

int main() {
    cin >> n >> randseed;
    int sum = 0;
    for (int i = n; i >= 1; i--) {
        a[i] = rnd() % 1000, b[i] = rnd() % 1000;
        if (sum + (a[i] - b[i]) < 0)
            swap(a[i], b[i]);
        sum += (a[i] - b[i]);
    }
    for (int i = 1; i <= n; i++) c[i] = a[i] - b[i];
    ll ans = 0, now = 0;
    for (int i = 1; i <= n; i++) {
        ans += now;
        if (c[i] < 0)  //有需求
            now += -c[i];
        else  //有供给
            now -= min(now, c[i]);
    }
    cout << ans << endl;
}