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HDU 4099 大数+Trie
Revenge of Fibonacci
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 204800/204800 K (Java/Others)
Total Submission(s): 3218 Accepted Submission(s): 821
Problem Description
The well-known Fibonacci sequence is defined as following:
Here we regard n as the index of the Fibonacci number F(n).
This sequence has been studied since the publication of Fibonacci‘s book Liber Abaci. So far, many properties of this sequence have been introduced.
You had been interested in this sequence, while after reading lots of papers about it. You think there’s no need to research in it anymore because of the lack of its unrevealed properties. Yesterday, you decided to study some other sequences like Lucas sequence instead.
Fibonacci came into your dream last night. “Stupid human beings. Lots of important properties of Fibonacci sequence have not been studied by anyone, for example, from the Fibonacci number 347746739…”
You woke up and couldn’t remember the whole number except the first few digits Fibonacci told you. You decided to write a program to find this number out in order to continue your research on Fibonacci sequence.
Here we regard n as the index of the Fibonacci number F(n).
This sequence has been studied since the publication of Fibonacci‘s book Liber Abaci. So far, many properties of this sequence have been introduced.
You had been interested in this sequence, while after reading lots of papers about it. You think there’s no need to research in it anymore because of the lack of its unrevealed properties. Yesterday, you decided to study some other sequences like Lucas sequence instead.
Fibonacci came into your dream last night. “Stupid human beings. Lots of important properties of Fibonacci sequence have not been studied by anyone, for example, from the Fibonacci number 347746739…”
You woke up and couldn’t remember the whole number except the first few digits Fibonacci told you. You decided to write a program to find this number out in order to continue your research on Fibonacci sequence.
Input
There
are multiple test cases. The first line of input contains a single
integer T denoting the number of test cases (T<=50000).
For each test case, there is a single line containing one non-empty string made up of at most 40 digits. And there won’t be any unnecessary leading zeroes.
For each test case, there is a single line containing one non-empty string made up of at most 40 digits. And there won’t be any unnecessary leading zeroes.
Output
For
each test case, output the smallest index of the smallest Fibonacci
number whose decimal notation begins with the given digits. If no
Fibonacci number with index smaller than 100000 satisfy that condition,
output -1 instead – you think what Fibonacci wants to told you beyonds
your ability.
Sample Input
15
1
12
123
1234
12345
9
98
987
9876
98765
89
32
51075176167176176176
347746739
5610
Sample Output
Case #1: 0
Case #2: 25
Case #3: 226
Case #4: 1628
Case #5: 49516
Case #6: 15
Case #7: 15
Case #8: 15
Case #9: 43764
Case #10: 49750
Case #11: 10
Case #12: 51
Case #13: -1
Case #14: 1233
Case #15: 22374
Source
2011 Asia Shanghai Regional Contest
debug三天竟是数据范围看错的惨案,将<=10w改成<10w后AC = =
还有一点关于大数模拟,一开始string一直错我以为是这个问题就改成手打的模拟大数加法,
关于这个char数组并不会自动初始化 假如 char s[35]; for(int i=0;i<10;++i) s[i]=i+‘0‘; 就会出现乱码导致我一直RE
真是醉了
#include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; #define ql(a) memset(a,0,sizeof(a)) #define LL long long const int UP=50; const int N=100000-1; struct node { int val; node *child[10]; node(){val=-1;for(int i=0;i<10;++i) child[i]=NULL;} }*root; void ins(char *s,int num) { node *p=root; int minn=min(40,(int)strlen(s)); for(int i=0;i<minn;++i){ int t=s[i]-‘0‘; if(p->child[t]==NULL){ p->child[t]=new node(); } p=p->child[t]; if(p->val<0) p->val=num; } } void init() { int f1[65],f2[65],f3[65],r=0; ql(f1),ql(f2),ql(f3); ins("1",0); f1[0]=f1[1]=f2[0]=f2[1]=1; for(int i=2;i<=N;++i){ql(f3);r=0; int ml=max(f1[0],f2[0]); for(int j=1;j<=ml;j++){ f3[j]=f1[j]+f2[j]+r; r=f3[j]/10; f3[j]%=10; if(j==ml&&r) ml++; }f3[0]=ml; char s[65]; ql(s);int l=0; for(int j=f3[0];j>=1;j--) s[l++]=f3[j]+‘0‘; for(int j=41;j<=f3[0];j++) s[j-1]=‘\0‘; ins(s,i); ql(f1); for(int j=0;j<=f2[0];j++) f1[j]=f2[j]; ql(f2); for(int j=0;j<=f3[0];j++) f2[j]=f3[j]; if(ml>55){ for(int j=1;j<f1[0];j++) f1[j]=f1[j+1]; f1[f1[0]--]=0; for(int j=1;j<f2[0];j++) f2[j]=f2[j+1]; f2[f2[0]--]=0; } } } int Find(char *s) { int len=strlen(s); if(!strcmp(s,"1")) {return 0;} node *p=root; for(int i=0;i<len;++i){ int t=s[i]-‘0‘; if(p->child[t]==NULL) return -1; p=p->child[t]; } return p->val; } int main() { int k,cas=0; char p[55]; root=new node(); init(); cin>>k; while(k--){ scanf("%s",p); printf("Case #%d: %d\n",++cas,Find(p)); } return 0; }
HDU 4099 大数+Trie
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