Vus the Cossack and Numbers CF-1186D（思维）

题意：

给一个浮点数序列$a$，其和为$0$。求一个序列$b$，满足$b[i]=ceil(a[i])$或者$b[i]=floor(a[i])$。

思路：

全部数字向下取整，赋值给$b$，统计$b$的和，若和不为$0$，则使原本不为整数的$a[i]$对应的$b[i]+1$，直到和为$0$。

代码：

//#include<bits/stdc++.h>
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>

#define ll long long
#define pll pair<ll,ll>
#define pii pair<int,int>
#define bug printf("*********
")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define ls root<<1
#define rs root<<1|1
#define pb push_back

using namespace std;
const int inf = 2e9 + 7;
const ll Inf = 1e18 + 7;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;

double a[maxn];
int b[maxn];

int main()
{
    int n;
    ll sum = 0;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i)
    {
        scanf("%lf", &a[i]);
        b[i] = floor(a[i]);
        sum += 1LL * b[i];
    }
    for (int i = 1; i <= n; ++i)
    {
        if (sum == 0)        break;
        if (a[i] - b[i] < 1e-7)    continue;
        b[i]++;
        sum++;
    }
    for (int i = 1; i <= n; ++i)    cout << b[i] << endl;
}