自从明明学了树的结构,就对奇怪的树产生了兴趣......给出标号为1到N的点,以及某些点最终的度数,允许在 任意两点间连线,可产生多少棵度数满足要求的树?
第一行为N(0 < N < = 1000), 接下来N行,第i+1行给出第i个节点的度数Di,如果对度数不要求,则输入-1
一个整数,表示不同的满足要求的树的个数,无解输出0
两棵树分别为1-2-3;1-3-2
purfer序列+高精度+质因数分解~
purfer序列真神奇~推荐看一下http://www.cnblogs.com/zhj5chengfeng/archive/2013/08/23/3278557.html~
注意像π一样的符号是从1乘到a[i]-1，不是直接乘a[i]-1，改过来以后代码就A了~
#include<cstdio> int n,x,cnt,tot,tot1,a[1001],num[169],ans[10001],len,c[169]={0,2,3,5,7,11,13,17,19,23,29,31,37,41, 43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151, 157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263, 269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383, 389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503, 509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641, 643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769, 773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911, 919,929,937,941,947,953,967,971,977,983,991,997}; void del(int u,int v) { for(int i=1;u>1 && i<=168;i++) while(!(u%c[i])) num[i]+=v,u/=c[i]; } void cheng(int v) { for(int i=1;i<=len;i++) ans[i]*=v; for(int i=1;i<=len;i++) if(ans[i]>=10) { ans[i+1]+=ans[i]/10;ans[i]%=10; if(i==len) len++; } } int main() { scanf("%d",&n); if(n==1) { scanf("%d",&x); if(!x) printf("1"); else printf("0"); return 0; } for(int i=1;i<=n;i++) { scanf("%d",&x); if(!x) { printf("0");return 0; } if(x!=-1) a[++cnt]=x,tot+=x-1; else tot1++; } if(tot>n-2) { printf("0");return 0; } for(int i=n-tot-1;i<=n-2;i++) del(i,1); for(int i=1;i<=cnt;i++) for(int j=1;j<a[i];j++) del(j,-1); del(n-cnt,n-2-tot); ans[1]=len=1; for(int i=1;i<=168;i++) while(num[i]--) cheng(c[i]); for(int i=len;i;i--) printf("%d",ans[i]); return 0; }
