CodeForces 1186D Vus the Cossack and Numbers(构造)

A - Vus the Cossack and Numbers

题意:

给你一些实数,可以向上取证也可以向下取整,要求取整之后序列的和为零

思路:

把能向下取整的数字全部向下取取整(已经是整数的数字是不能向下取整的),然后看总和比零小多少,再把一些数字改成向上取整的

 1 #include<cstdio>
 2 #include<string.h>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<iostream>
 6 #include<vector>
 7 #include<queue>
 8 #include<set>
 9 #include<map>
10 #include<cctype>
11 #define ios ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
12 #define mem(a,x) memset(a,x,sizeof(a))
13 #define lson rt<<1,l,mid
14 #define rson rt<<1|1,mid + 1,r
15 #define P pair<int,int>
16 #define ull unsigned long long
17 using namespace std;
18 typedef long long ll;
19 const int maxn = 1e6 + 10;
20 const ll mod = 998244353;
21 const int inf = 0x3f3f3f3f;
22 const long long INF = 0x3f3f3f3f3f3f3f3f;
23 const double eps = 1e-7;
24 inline ll read()
25 {
26     ll X = 0, w = 0; char ch = 0;
27     while (!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
28     while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
29     return w ? -X : X;
30 }
31 
32 double arr[maxn];
33 int brr[maxn];
34 int main()
35 {
36     int n;
37     while (~scanf("%d", &n))
38     {
39         int sum = 0;
40         for (int i = 1; i <= n; ++i)
41         {
42             scanf("%lf", &arr[i]);
43             if (arr[i] < 0)
44             {
45                 brr[i] = (int)(arr[i] - 0.99999999);
46             }
47             else
48             {
49                 brr[i] = (int)arr[i];
50             }
51             sum += brr[i];
52         }
53         for (int i = 1; i <= n; ++i)
54         {
55             if (sum == 0) break;
56             if (arr[i] - (int)arr[i] == 0) continue;
57             if (arr[i] < 0) brr[i] = (int)arr[i], sum++;
58             else
59                 brr[i] = (int)(arr[i] + 0.99999999) , sum++;
60         }
61         for (int i = 1; i <= n; ++i)
62         {
63             cout << brr[i] << endl;
64         }
65     }
66     return 0;
67 }
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posted @ 2020-04-13 21:36  当然是斗笠呀  阅读(179)  评论(0编辑  收藏  举报