CF1186 D. Vus the Cossack and Numbers


给定\(n\)个浮点数\(a_i\),满足\(\sum\limits_{i=1}^na_i=0\),令\(b_i=\lfloor a_i\rfloor\)\(b_i=\lceil a_i\rceil\),求构造一组\(b_i\),满足\(\sum\limits_{i=1}^nb_i=0\)

首先令\(b_i=\lfloor a_i\rfloor\),这样\(\sum\limits_{i=1}^nb_i<0\),我们再根据\(|\sum\limits_{i=1}^nb_i|\)的值,将一些\(b_i\)改为\(b_i+1\)即可

注意存在\(\lfloor a_i\rfloor=\lceil a_i\rceil\)的情况,这样\(b_i\)是不能改成\(b_i+1\)

/*program from Wolfycz*/
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
template<typename T>inline T frd(T x){
	int f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')    f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
template<typename T>inline T read(T x){
	int f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
const int N=1e5;
const double eps=1e-8;
double A[N+10];
int B[N+10];
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	int n=read(0),All=0;
	for (int i=1;i<=n;i++){
	for (int i=1;All;i++){
		if (abs(A[i]-B[i])<=eps)	continue;
	for (int i=1;i<=n;i++)	printf("%d\n",B[i]);
	return 0;
posted @ 2021-07-08 10:43  Wolfycz  阅读(18)  评论(0编辑  收藏  举报