[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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