C++ C+中的分治算法+;
在一些在线法官身上有一个问题,我不知道如何被接受 问题是第一行包含两个数字C++ C+中的分治算法+;,c++,algorithm,divide-and-conquer,C++,Algorithm,Divide And Conquer,在一些在线法官身上有一个问题,我不知道如何被接受 问题是第一行包含两个数字 N (0 < N < 2^18) M (0 < M < 2^20) 问题是有多少X满足以下条件: M = floor(X/a1) + floor(X/a2) + ... + floor(X/an) 我天真的解决方案: #include<bits/stdc++.h> using namespace std; long long n,m,i,j,haha,sum; int main
N (0 < N < 2^18)
M (0 < M < 2^20)
问题是有多少X
满足以下条件:
M = floor(X/a1) + floor(X/a2) + ... + floor(X/an)
我天真的解决方案:
#include<bits/stdc++.h>
using namespace std;
long long n,m,i,j,haha,sum;
int main()
{
cin >> n >> m;
haha = 0;
long long ar[n+5];
for(i = 0; i < n; i++) cin >> ar[i];
sort(ar,ar+n);
for(i = ar[0]+1; i < m*ar[0]; i++){
sum = 0;
for (j = 0; j < n; j++) sum += i/ar[j];
if (sum == m) haha += 1;
else if (sum >= m) break;
}
cout << haha << endl;
}
#包括
使用名称空间std;
龙龙n,m,i,j,哈哈,和;
int main()
{
cin>>n>>m;
哈哈=0;
长ar[n+5];
对于(i=0;i>ar[i];
排序(ar,ar+n);
对于(i=ar[0]+1;i=m)中断,则为else;
}
无法更新
使用二进制搜索方法,下面是我的新代码:
// compute X/ai sum
long long summarize(long long ar[], long long n, long long X)
{
long long sum = 0;
for (long long i = 0; i < n; i++)
{
sum += X/ar[i];
}
return sum;
}
bool get_range(long long ar[], int n, int m, pair<long long, long long>& range)
{
long long sum = 0;
long long x;
// reduce range
while (range.first < range.second)
{
x = (range.first + range.second) / 2;
sum = summarize(ar, n, x);
if (sum < m)
{
range.first = x + 1;
}
else if (sum > m)
{
range.second = x;
}
else if (x == range.first)
{
return true; // single element
}
else
{
break;
}
}
if (sum != m)
{
return false;
}
// check surroundings for lower / upper bound.
sum = summarize(ar, n, range.first);
if (sum != m)
{
auto r1 = make_pair(range.first + 1, x);
if (get_range(ar, n, m, r1))
{
range.first = r1.first;
}
else
{
range.first = x;
}
}
sum = summarize(ar, n, range.second - 1);
if (sum != m)
{
auto r2 = make_pair(x + 1, range.second - 1);
if (get_range(ar, n, m, r2))
{
range.second = r2.second;
}
else
{
range.second = x + 1;
}
}
return true;
}
int main()
{
int n, m;
cin >> n >> m;
long long *ar = new long long[n];
long long ar_min = LLONG_MAX;
for(long long i = 0; i < n; i++)
{
cin >> ar[i];
ar_min = min(ar[i], ar_min);
}
// initial range of possible X values
auto range = make_pair(m / (ar_min * n), m * ar_min);
if (get_range(ar, n, m, range))
{
cout << (range.second - range.first) << endl;
}
else
{
cout << 0 << endl;
}
}
对于每个子项floor(X/a1)
,都有floor(X1/ai)n>>m;
long-long*ar=新的long-long[n];
long long ar_最小值=long_最大值;
for(长i=0;i>ar[i];
ar_min=min(ar[i],ar_min);
}
//最低可能k
长k=m/(ar_min*n);
//获取可能的X值范围的值k
对于(;km)
{
打破
}
}
长X_min=k*ar_min,X_max=(k+1)*ar_min;
长结果=0;
//计算可能的X值
用于(长x=x_最小值;xm)
{
打破
}
}
cout我相信这个问题的预期解决方案是二进制搜索
定义f(x)=sum\u i f(x/a\u i)
。在不丧失一般性的情况下,假设a_i
以递增顺序给出
显然,
f(0)=0
f(M*a_1)≥ M
f(x)≥ f(y)如果x≥y
因此,您可以执行二进制搜索以查找x的最低值,从而f(x)=M
,并将start=0
和end=M*a_1
作为二进制搜索的初始限制
要找到x的上限,请执行另一个二进制搜索,或只是循环遍历数组中的所有值,以找到最小的y
,这样floor(y/ai)>floor(x/ai)
对于某些i
使用两个二进制搜索(分别用于下界和上界)和此代码(最终)被接受:
#include<bits/stdc++.h>
using namespace std;
long long n,m,i,l,r,mid1,mid2,ans,tmp,cnt,haha,k;
long long ar[26214400];
long long func(long long x){
haha = 0;
for (k = 0; k < n; k++) haha += x/ar[k];
return haha;
}
int main()
{
cin >> n >> m;
for(i = 0; i < n; i++) cin >> ar[i];
sort(ar,ar+n);
l = ar[0];
r = ar[0]*m;
mid1 = (l+r)/2;
tmp = func(mid1);
while (l < r){
mid1 = (l+r)/2;
tmp = func(mid1);
if (tmp < m) l = mid1+1;
else if (tmp > m) r = mid1-1;
else r = mid1-1;
}
mid1 = l; //lower bound
l = ar[0];
r = ar[0]*m;
mid2 = (l+r)/2;
tmp = func(mid2);
while (l < r){
mid2 = (l+r)/2;
tmp = func(mid2);
if (tmp < m) l = mid2+1;
else if (tmp > m) r = mid2-1;
else l = mid2+1;
}
mid2 = r; //upper bound
while (mid1 <= mid2 and func(mid1) != m) mid1 += 1;
while (mid2 >= mid1 and func(mid2) != m) mid2 -= 1;
ans = mid2-mid1+1;
cout << ans << endl;
}
#包括
使用名称空间std;
长长n,m,i,l,r,mid1,mid2,ans,tmp,cnt,哈哈,k;
long-long ar[26214400];
长函数(长x){
哈哈=0;
对于(k=0;k>n>>m;
对于(i=0;i>ar[i];
排序(ar,ar+n);
l=ar[0];
r=ar[0]*m;
mid1=(l+r)/2;
tmp=func(mid1);
while(lm)r=mid1-1;
否则r=mid1-1;
}
mid1=l;//下限
l=ar[0];
r=ar[0]*m;
mid2=(l+r)/2;
tmp=func(mid2);
while(lm)r=mid2-1;
否则l=mid2+1;
}
mid2=r;//上界
而(mid1=mid1和func(mid2)!=m)mid2-=1;
ans=mid2-mid1+1;
你能提供问题链接吗?@User\u Targaryen这个问题是用我的母语写的,来自当地的在线法官。证明这个链接不会有帮助。如果你真的想知道的话。@Aldihilman如果问题应该由D&C解决,我想这里唯一的选择是使用二进制搜索。谓词是“X使这个表达式等于M吗?",如果否,则沿已排序的X值向左或向右移动,如果是,则停止并检查X@grek40不起作用。例如,N
=3和M
=10。我们得到了a1
=1、a2
=1和a3
=1。我们不可能得到X
。然后展示您在二进制搜索方面的尝试。I讨论线性搜索毫无意义。不错的算法。但是你的算法比我最新的使用二进制搜索的算法慢。是的,你的算法也没有通过时间限制,它甚至得到了更低的分数。是的,后来我突然想到了“除法与合并”这个标题是密钥…正在进行更新,但尚未完成。是的。您的二进制搜索算法的分数与我的二进制搜索版本代码的分数相同。而且,您猜,它没有超过时间限制。@Aldihilman您是否有任何时间关键的示例测试用例或所有测试都是黑盒?很遗憾,黑盒。请共享您的二进制搜索代码,pe可以改进的rhaps。问题不在于二进制搜索,而在于二进制搜索后的while循环。这需要花费太长的时间。您应该使用二进制搜索来查找mid的最低值,而不是在第一个实例中中断。并使用另一个二进制搜索来查找第一个值,其中提供了更大的func。
// compute X/ai sum
long long summarize(long long ar[], long long n, long long X)
{
long long sum = 0;
for (long long i = 0; i < n; i++)
{
sum += X/ar[i];
}
return sum;
}
bool get_range(long long ar[], int n, int m, pair<long long, long long>& range)
{
long long sum = 0;
long long x;
// reduce range
while (range.first < range.second)
{
x = (range.first + range.second) / 2;
sum = summarize(ar, n, x);
if (sum < m)
{
range.first = x + 1;
}
else if (sum > m)
{
range.second = x;
}
else if (x == range.first)
{
return true; // single element
}
else
{
break;
}
}
if (sum != m)
{
return false;
}
// check surroundings for lower / upper bound.
sum = summarize(ar, n, range.first);
if (sum != m)
{
auto r1 = make_pair(range.first + 1, x);
if (get_range(ar, n, m, r1))
{
range.first = r1.first;
}
else
{
range.first = x;
}
}
sum = summarize(ar, n, range.second - 1);
if (sum != m)
{
auto r2 = make_pair(x + 1, range.second - 1);
if (get_range(ar, n, m, r2))
{
range.second = r2.second;
}
else
{
range.second = x + 1;
}
}
return true;
}
int main()
{
int n, m;
cin >> n >> m;
long long *ar = new long long[n];
long long ar_min = LLONG_MAX;
for(long long i = 0; i < n; i++)
{
cin >> ar[i];
ar_min = min(ar[i], ar_min);
}
// initial range of possible X values
auto range = make_pair(m / (ar_min * n), m * ar_min);
if (get_range(ar, n, m, range))
{
cout << (range.second - range.first) << endl;
}
else
{
cout << 0 << endl;
}
}
M = floor(X/a1) + floor(X/a2) + ... + floor(X/an)
// compute X/ai sum
long long summarize(long long ar[], long long n, long long X)
{
long long sum = 0;
for (long long i = 0; i < n; i++)
{
sum += X/ar[i];
}
return sum;
}
int main()
{
int n, m;
cin >> n >> m;
long long *ar = new long long[n];
long long ar_min = LLONG_MAX;
for(long long i = 0; i < n; i++)
{
cin >> ar[i];
ar_min = min(ar[i], ar_min);
}
// lowest possible k
long long k = m / (ar_min * n);
// get the value k for a possible range of X values
for (; k <= m; k++)
{
auto x = ar_min * (k + 1);
long long sum = summarize(ar, n, x);
if (sum > m)
{
break;
}
}
long long X_min = k * ar_min, X_max = (k + 1) * ar_min;
long long result = 0;
// count possible X values
for (long long x = X_min; x < X_max; x++)
{
long long sum = summarize(ar, n, x);
if (sum == m)
{
++result;
}
else if (sum > m)
{
break;
}
}
cout << result << endl;
}
#include<bits/stdc++.h>
using namespace std;
long long n,m,i,l,r,mid1,mid2,ans,tmp,cnt,haha,k;
long long ar[26214400];
long long func(long long x){
haha = 0;
for (k = 0; k < n; k++) haha += x/ar[k];
return haha;
}
int main()
{
cin >> n >> m;
for(i = 0; i < n; i++) cin >> ar[i];
sort(ar,ar+n);
l = ar[0];
r = ar[0]*m;
mid1 = (l+r)/2;
tmp = func(mid1);
while (l < r){
mid1 = (l+r)/2;
tmp = func(mid1);
if (tmp < m) l = mid1+1;
else if (tmp > m) r = mid1-1;
else r = mid1-1;
}
mid1 = l; //lower bound
l = ar[0];
r = ar[0]*m;
mid2 = (l+r)/2;
tmp = func(mid2);
while (l < r){
mid2 = (l+r)/2;
tmp = func(mid2);
if (tmp < m) l = mid2+1;
else if (tmp > m) r = mid2-1;
else l = mid2+1;
}
mid2 = r; //upper bound
while (mid1 <= mid2 and func(mid1) != m) mid1 += 1;
while (mid2 >= mid1 and func(mid2) != m) mid2 -= 1;
ans = mid2-mid1+1;
cout << ans << endl;
}