描述
http://poj.org/problem?id=2100
求连续平方和=n的序列个数,并输出序列.
Graveyard Design
| Time Limit: 10000MS | Memory Limit: 64000K | |
| Total Submissions: 5987 | Accepted: 1416 | |
| Case Time Limit: 2000MS | ||
Description
King George has recently decided that he would like to have a new design for the royal graveyard. The graveyard must consist of several sections, each of which must be a square of graves. All sections must have different number of graves.
After a consultation with his astrologer, King George decided that the lengths of section sides must be a sequence of successive positive integer numbers. A section with side length s contains s2 graves. George has estimated the total number of graves that will be located on the graveyard and now wants to know all possible graveyard designs satisfying the condition. You were asked to find them.
After a consultation with his astrologer, King George decided that the lengths of section sides must be a sequence of successive positive integer numbers. A section with side length s contains s2 graves. George has estimated the total number of graves that will be located on the graveyard and now wants to know all possible graveyard designs satisfying the condition. You were asked to find them.
Input
Input file contains n --- the number of graves to be located in the graveyard (1 <= n <= 1014 ).
Output
On
the first line of the output file print k --- the number of possible
graveyard designs. Next k lines must contain the descriptions of the
graveyards. Each line must start with l --- the number of sections in
the corresponding graveyard, followed by l integers --- the lengths of
section sides (successive positive integer numbers). Output line's in
descending order of l.
Sample Input
2030
Sample Output
2 4 21 22 23 24 3 25 26 27
Source
Northeastern Europe 2004, Northern Subregion
分析
直接尺取.
注意:
1.如果要用开根计算的话要写成 " ll ub=(ll)sqrt(n*1.0); "写成 " ll ub=sqrt(n); "会CE.
所以干脆写成 " r*r<=n "
1 #include<cstdio> 2 #include<queue> 3 #define ll long long 4 using std :: queue; 5 6 struct node 7 { 8 ll len,fst; 9 node() {} 10 node(ll a,ll b) : len(a),fst(b) {} 11 }; 12 queue <node> q; 13 ll n; 14 15 inline ll val(ll x) { return x*x; } 16 17 int main() 18 { 19 #ifndef ONLINE_JUDGE 20 freopen("grave.in","r",stdin); 21 freopen("grave.out","w",stdout); 22 #endif 23 scanf("%lld",&n); 24 ll l=1,r=0,k=0,len=0; 25 ll sum=0; 26 while(val(r)<=n) 27 { 28 if(sum<n) 29 { 30 sum+=val(++r); 31 len++; 32 } 33 else if(sum>n) 34 { 35 sum-=val(l++); 36 len--; 37 } 38 else 39 { 40 q.push(node(len,l)); 41 k++; 42 sum-=val(l++); 43 len--; 44 } 45 } 46 printf("%lld",k); 47 while(!q.empty()) 48 { 49 node t=q.front(); q.pop(); 50 ll len=t.len,fst=t.fst; 51 printf(" %lld ",len); 52 for(ll i=0;i<len;i++) printf("%lld ",fst+i); 53 } 54 #ifndef ONLINE_JUDGE 55 fclose(stdin); 56 fclose(stdout); 57 #endif 58 return 0; 59 }