POJ 2042

打表解之

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 1 #include<cstdio>

 2 #include<cstring>

 3 #include<algorithm>

 4 #include<cmath>

 5 using namespace std;

 6 const int N=(1<<15)+1;

 7 int po[200],ans[N+2];

 8 int main()

 9 {

10     for(int i=1;i<200;po[i]=i*i,i++);

11     memset(ans,0,sizeof(ans));

12     for(int i1=1,tp1;po[i1]<=N;i1++)

13     {

14         ans[tp1=po[i1]]++;

15         for(int i2=i1,tp2;tp1+po[i2]<=N;i2++)

16         {

17             ans[tp2=tp1+po[i2]]++;

18             for(int i3=i2,tp3;tp2+po[i3]<=N;i3++)

19             {

20                 ans[tp3=tp2+po[i3]]++;

21                 for(int i4=i3;tp3+po[i4]<=N;i4++)

22                     ans[tp3+po[i4]]++;

23             }

24         }

25     }

26     int n;

27     while(scanf("%d",&n)&&n)

28         printf("%d\n",ans[n]);

29     return 0;

30 }