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 }