codeforces A. Jeff and Rounding (数学公式+贪心)

题目链接：http://codeforces.com/contest/351/problem/A

 

算法思路：2n个整数，一半向上取整，一半向下。我们设2n个整数的小数部分和为sum.

ans = |A - B|;

sum = A +(n-b)-B;

所以ans = |sum - (n-b) |; 只有b未知，只需要枚举一下b就得到答案。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;

const int maxn = 2005;
const double eps = 1e-12;
int dcmp(double x){
    if(fabs(x) < eps) return 0;
    else              return x < 0 ? -1 : 1;
}
int main()
{
    //freopen("E:\\acm\\input.txt","r",stdin);

    double sum = 0;
    int n;
    int num = 0;
    cin>>n;
    for(int i=1;i<=2*n;i++)
    {
        double a;
        scanf("%lf",&a);
        if(dcmp(floor(a)-a) == 0) num++;
        sum += a - floor(a);
    }
    double ans = 1e10;
    for(int i=max(num-n,0);i<=min(n,num);i++)
    {
        ans = min(ans,fabs(sum-(n-i)));
    }
    printf("%.3lf\n",ans);
}