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CF1186D - Vus the Cossack and Numbers

作者:互联网

传送门:Problem - 1186D - Codeforces

输入的数要么向上取整,要么向下取整,那么设ans为负数的绝对值的和,sum为正数的和,对单一的负数向下取整于负数而言相当于ans++,同理正数向上取整sum++。

这里我们设ans1为ans的最小值即负数全部向上取整,ans2为ans最大值,sum1,sum2同理

由于题目保证答案存在,故ans1必定小于sum2,我们只要通过改变取整方式,让它们“会合”即可使得新序列的和为0

我们遍历序列,遇见负数改变ans1的值,遇见正数改变sum2的值,ans1++即输出负数向下取整,sum2--即输出正数向下取整,一旦ans1==ans2时负数变为向上取整,ans1停止变化,sum1==sum2时正数变为向上取整,ans1==sum2时同理。

 

此时输出的序列符合条件

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
#define ll long long
using namespace std;


ll n,m;

double a[500010];
ll sum1,sum2;
ll ans1,ans2;
int main(){
    cin>>n;
    for(ll i=1;i<=n;i++)
    {
        ll flag=0;
        double x;
        cin>>x;
        a[i]=x;
        if(x<0)flag=1;
        x=fabs(x);
        ll y;
        y=x*1;
        if(x==y)
        {
            if(flag){
                ans1+=y;
                ans2+=y;
            }
            else{
                sum1+=y;
                sum2+=y;
            }
        }
        else{
            if(flag)
            {
                ans1+=y;
                ans2+=(y+1);
            }
            else{
                sum1+=y;
                sum2+=(y+1);
            }
        }
    }
    //cout<<ans2<<" "<<sum1<<endl;
    for(ll i=1;i<=n;i++)
    {
        if(ans2!=sum1)
        {
            ll y=a[i]*1;
            if(y==a[i])
            {
                cout<<y<<endl;
                continue;
            }
            if(a[i]<0)
            {
                //ll y=a[i]*1;
                cout<<y<<endl;
                ans2--;
            }
            else{
                //ll y=a[i]*1;
                cout<<y+1<<endl;
                sum1++;
            }
        }
        else {
            ll y=a[i]*1;
            if(y==a[i])
            {
                cout<<y<<endl;
                continue;
            }
            if(a[i]<0)
            {
                //ll y=a[i]*1;
                cout<<y-1<<endl;
            }
            else{
                //ll y=a[i]*1;
                cout<<y<<endl;
            }
        }
    }
    return 0;
}

 

标签:ans1,ll,sum2,负数,取整,Vus,CF1186D,include,Cossack
来源: https://www.cnblogs.com/lemonGJacky/p/15025714.html