解法一：前缀和+二分查找

先求前缀和，遍历每个位置作为起点，用二分查找找到最小费用的终点，不断更新最小费用及其方案队列。

#include <bits/stdc++.h>
using namespace std;
int N, M;
int sum[100005];
int binSearch(int i, int &right) {
    right = N;
    int left = i;
    int mid, val;
    while (left < right) {
        mid = (left + right) / 2;
        val = sum[mid] - sum[i - 1];
        if (val >= M) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return sum[right] - sum[i - 1];
}
vector<pair<int, int>> result;
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> N >> M;
    sum[0] = 0;
    for (int i = 1; i <= N; ++i) {
        cin >> sum[i];
        sum[i] += sum[i - 1];
    }
    int minCost = sum[N];
    for (int i = 1; i <= N; ++i) {
        int j;
        int cost = binSearch(i, j);
        if (cost >= M && cost <= minCost) {
            if (cost < minCost) {
                minCost = cost;
                result.clear();
            }
            result.emplace_back(make_pair(i, j));
        }
    }
    for (auto i:result) {
        cout << i.first << "-" << i.second << "\n";
    }
}