如题，复杂度是 $O(nlogn)$ ，方法是用前缀和判断二分条件然后用 ST 表求静态 RMQ

50pts WA

#include <bits/stdc++.h>
using namespace std;

const int NR = 1e5 + 5;
const int LOGN = 21;
const int INF = 2147483647;

int n, m;
int qwq, f[NR][LOGN], a[NR], s[NR];

int find(int x, int r) {
	int l = 1;
	int mid = (l + r) >> 1;
	while (l <= r) {
		mid = (l + r) >> 1;
		if (s[r] - s[mid-1] == m) {
			return mid;
		}
		if (s[r] - s[mid-1] > m) {
			l = mid + 1;
		}
		if (s[r] - s[mid-1] < m) {
			r = mid - 1;
		}
	}
	return mid;
}

int query(int l, int r) {
	int s = log2(r - l + 1);
    return max(f[l][s], f[r-(1<<s)+1][s]);
}

int main() {
	s[0] = 0;
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf("%d%d", &qwq, &a[i]);
		s[i] = s[i-1] + qwq;
		f[i][0] = a[i];
	}
	
	for (int j = 1; j <= LOGN; j++) {
		for (int i = 1; i + (1 << j) - 1 <= n; i++) {
			f[i][j] = max(f[i][j-1], f[i+(1<<(j-1))][j-1]);
		}
	}
	
	int ans = INF;
	int l = 1;
	for (int r = 1; r <= n; r++) {
		if (s[r] - s[l-1] < m) continue ;
		int l = find(s[r] - m, r);
		ans = min(ans, query(l, r));
	}
	cout << ans << endl;
	
	
	return 0;
}