[暑假集训--数论]poj2034 Anti-prime Sequences
We can extend the definition by defining a degree danti-prime sequence as one where all consecutive subsequences of length 2,3,...,d sum to a composite number. The sequence above is a degree 2 anti-prime sequence, but not a degree 3, since the subsequence 5, 4, 2 sums to 11. The lexicographically .rst degree 3 anti-prime sequence for these numbers is 1,3,5,4,6,2,10,8,7,9.
Input
Output
No anti-prime sequence exists.
Sample Input
1 10 2
1 10 3
1 10 5
40 60 7
0 0 0
Sample Output
1,3,5,4,2,6,9,7,8,10
1,3,5,4,6,2,10,8,7,9
No anti-prime sequence exists.
40,41,43,42,44,46,45,47,48,50,55,53,52,60,56,49,51,59,58,57,54
需要一个 l 到 r 的排列,使得任意一个长度是2,3,...,k的子串当中子串和都不是质数
直接搜就是了
#include<cstdio>
#include<iostream>
#include<cstring>
#define LL long long
using namespace std;
inline LL read()
{
LL x=,f=;char ch=getchar();
while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
return x*f;
}
int l,r,d,haveans=;
bool mk[];
bool mrk[];
int s[];
int p[],len;
inline void getp()
{
for (int i=;i<=;i++)
{
if (!mk[i])
{
p[++len]=i;
for (int j=*i;j<=;j+=i)mk[j]=;
}
}
}
inline void dfs(int now)
{
if (now==r+)
{
for (int i=l;i<r;i++)printf("%d,",s[i]);
printf("%d\n",s[r]);
haveans=;
return;
}
for (int i=l;i<=r;i++)
{
if (mrk[i])continue;
int sum=i,mrk2=;
for (int j=now-;j>=max(l,now-d+);j--)
{
sum+=s[j];
if(!mk[sum]){mrk2=;break;}
}
if (!mrk2)continue;
mrk[i]=;
s[now]=i;
dfs(now+);
if (haveans)return;
s[now]=;
mrk[i]=;
}
}
int main()
{
getp();
while (~scanf("%d%d%d",&l,&r,&d)&&l+r+d)
{
memset(mrk,,sizeof(mrk));
haveans=;
dfs(l);
if (!haveans)puts("No anti-prime sequence exists.");
}
}
poj 2034
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