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 }