poj 2100 尺取法 一个数字拆成连续数字平方和

题意：将一个数拆成若干个连续数字的平方和。

用尺取法枚举区间，复杂度为O(n)，时限10s，3s多ac。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <cmath>
 5 using namespace std;
 6
 7 const int N = 100;
 8
 9 struct Node
10 {
11     int from, to;
12 } node[N];
13
14 void solve( long long n )
15 {
16     int u = sqrt( n * 1.0 ), cnt = 0, q = 1;
17     long long sum = 0;
18     for ( int i = 1; i <= u; i++ )
19     {
20         sum += ( long long ) i * i;
21         while ( sum > n )
22         {
23             sum -= ( long long ) q * q;
24             q++;
25         }
26         if ( sum == n )
27         {
28             node[cnt].from = q;
29             node[cnt].to = i;
30             cnt++;
31         }
32     }
33     printf("%d\n", cnt);
34     for ( int i = 0; i < cnt; i++ )
35     {
36         printf("%d", node[i].to - node[i].from + 1);
37         for ( int j = node[i].from; j <= node[i].to; j++ )
38         {
39             printf(" %d", j);
40         }
41         puts("");
42     }
43 }
44
45 int main ()
46 {
47     long long n;
48     while ( scanf("%lld", &n) != EOF )
49     {
50         solve(n);
51     }
52     return 0;
53 }