递归函数

    int Top_Down_Rod(int p[],int n){
        int r=0;
        int i;
        if(n==0)return 0;
        for (i=1;i<=n;i++){
            int tmp=p[i]+Top_Down_Rod(p,n-i);
            count=count+1;
            System.out.println("第"+count+"次"+(n));
            r=(r>=tmp)?r:tmp;
        }
        return r;
    }

T(n)=T(n-1)+...T(0);

T(n)=2的n次方减一