CF4 A与B

最新推荐文章于 2024-03-28 18:43:25 发布

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最新推荐文章于 2024-03-28 18:43:25 发布

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本文链接：https://blog.youkuaiyun.com/sdjzujxc/article/details/8246452

本博客提供了一个简单的C++代码示例，用于判断一个数是否能被拆分为两个偶数的和，并讨论了如何合理安排生物考试前的复习计划，确保总复习时间符合父母设定的限制。

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A简单题，判断一个数能否分成两个偶数的和。

#include<iostream>
using namespace std;
int main()
{
    int n;
    while(cin>>n)
    {
        if(n%2==0&&n!=2)
        cout<<"YES"<<endl;
        else 
        cout<<"NO"<<endl;
    }
    return 0;
}

B：

B - Before an Exam

Crawling in process... Crawling failed Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 4B

Description

Tomorrow Peter has a Biology exam. He does not like this subject much, but d days ago he learnt that he would have to take this exam. Peter's strict parents made him prepare for the exam immediately, for this purpose he has to study not less than minTime_i and not more than maxTime_i hours per each i-th day. Moreover, they warned Peter that a day before the exam they would check how he has followed their instructions.

So, today is the day when Peter's parents ask him to show the timetable of his preparatory studies. But the boy has counted only the sum of hours sumTime spent him on preparation, and now he wants to know if he can show his parents a timetable sсhedule with d numbers, where each number sсhedule_i stands for the time in hours spent by Peter each i-th day on biology studies, and satisfying the limitations imposed by his parents, and at the same time the sum total of all schedule_i should equal to sumTime.

Input

The first input line contains two integer numbers d, sumTime (1 ≤ d ≤ 30, 0 ≤ sumTime ≤ 240) — the amount of days, during which Peter studied, and the total amount of hours, spent on preparation. Each of the following d lines contains two integer numbers minTime_i, maxTime_i (0 ≤ minTime_i ≤ maxTime_i ≤ 8), separated by a space — minimum and maximum amount of hours that Peter could spent in the i-th day.

Output

In the first line print YES, and in the second line print d numbers (separated by a space), each of the numbers — amount of hours, spent by Peter on preparation in the corresponding day, if he followed his parents' instructions; or print NO in the unique line. If there are many solutions, print any of them.

Sample Input

Input

1 48
5 7

Output

NO

Input

2 5
0 1
3 5

Output

YES
1 4

#include<iostream>
using namespace std;
struct node
{
    int mint;
    int maxt;
}p[35];
int tmp[35];
int n,sum;
int main()
{
    while(cin>>n>>sum)
    {
        int sum1=0;
        int sum2=0;
        int sum3=0;
        int k;
        for(int i=0;i<n;i++)
        {
            cin>>p[i].mint>>p[i].maxt;
            sum1+=p[i].mint;
            sum2+=p[i].maxt;
            tmp[i]=p[i].mint;
        }
        if(sum2<sum||sum<sum1)
        cout<<"NO"<<endl;
        else
        {
            cout<<"YES"<<endl;
            sum-=sum1;
            for(int i=0;i<n;i++)
            {
                int ans=p[i].maxt-p[i].mint;
                if(sum>ans)
                {
                   tmp[i]+=ans;
                   sum-=ans;
                }
                else
                {
                    tmp[i]+=sum;
                    break;
                }
            }
            for(int i=0;i<n-1;i++)
            cout<<tmp[i]<<" ";
            cout<<tmp[n-1]<<endl;
        }
    }
    return 0;
}
/*
2 5
0 1
3 5
*/