hdu 2510 符号三角形 回溯

2^n的回溯算法，n<=24,打表即可。

#include<iostream>#include<cstring>#include<cstdio>using namespace std;#define MAX 25//ashione 2012-4-19//打表，不打表会超时，因为回溯算法的时间复杂度是很高的。注释掉的代码是打表的。int q[MAX][MAX];int readc[MAX]={0,0,0,4,6,0,0,12,40,0,0,171,410,0,0,1896,5160,0,0,32757,59984,0,0,431095,822229};/*int sum,count;void make(int t,int n){         if(count>(n+1)*(n)/4 || (t-1)*t/2 - count>(n+1)*n/4)                return ;         if(t>n){                 sum++;//               cout<<"hello"<<endl;                 return ;         }         for(int i=0;i<2;i++){                 q[1][t] = i;                 count+=i;                 for(int j=2;j<=t;j++){                                q[j][t-j+1] = q[j-1][t-j+1]^q[j-1][t-j+2];                                count+=q[j][t-j+1];                 }                 make(t+1,n);                 for(int j=1;j<=t;j++)                         count-=q[j][t-j+1];         }}*/int main(){        int n;/*        for(int i=1;i<=24;i++){                count = sum= 0;                memset(q,0,sizeof(q));                if((i*(i+1)/2)%2 == 0)                        make(1,i);                        readc[i] =sum;                cout<<readc[i]<<",";        }        cout<<endl;*/        while(cin>>n && n){                cout<<n<<" "<<readc[n]<<endl;        }        return 0;}