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牛客 24086 Haybale Feast

阚元白
2023-12-01

牛客 24086 Haybale Feast

题目链接

题目描述

Farmer John is preparing a delicious meal for his cows! In his barn, he has N haybales ( 1 ≤ N ≤ 100 , 000 ) (1≤N≤100,000) (1N100,000). The iith haybale has a certain flavor F i ( 1 ≤ F i ≤ 1 0 9 ) F_i (1≤F_i≤10^9 ) Fi(1Fi109) and a certain spiciness S i ​ ( 1 ≤ S i ≤ 1 0 9 ) S_i ​ (1 \le S_i \le 10^9 ) Si(1Si109).

The meal will consist of a single course, being a contiguous interval containing one or more consecutive haybales (Farmer John cannot change the order of the haybales). The total flavor of the meal is the sum of the flavors in the interval. The spiciness of the meal is the maximum spiciness of all haybales in the interval.

Farmer John would like to determine the minimum spiciness his single-course meal could achieve, given that it must have a total flavor of at least M ( 1 ≤ M ≤ 1 0 18 ) M (1 \le M \le 10^{18} ) M(1M1018).

输入描述:

The first line contains the integers N and M, the number of haybales and the minimum total flavor the meal must have, respectively. The next N lines describe the N haybales with two integers per line, first the flavor F and then the spiciness S.

输出描述:

Please output the minimum spiciness in a single course meal that satisfies the minimum flavor requirement. There will always be at least one single-course meal that satisfies the flavor requirement.

示例1

输入

5 10
4 10
6 15
3 5
4 9
3 6

输出

9

简单二分:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e5+5;
int n;
ll m,a[N],b[N];
bool check(ll x){
    ll sum=0;
    for(int i=0;i<n;i++){
        if(b[i]>x){
            sum=0;
            continue;
        }
        sum+=a[i];
        if(sum>=m) return 1;
    }
    return 0;
}

int main()
{
    cin>>n>>m;
    for(int i=0;i<n;i++) cin>>a[i]>>b[i];
    ll l=0,r=1e18;
    while(l<=r){
        ll mid=l+r>>1;
        if(check(mid)) r=mid-1;
        else l=mid+1;
    }
    cout<<l<<endl;
	return 0;
}
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