HDU 4099 大數+Trie
阿新 • • 發佈:2017-07-25
pri struct member 亂碼 stream prop reg reading cstring
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.
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
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
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.
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.
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三天竟是數據範圍看錯的慘案,將<=10w改成<10w後AC = = 還有一點關於大數模擬,一開始string一直錯我以為是這個問題就改成手打的模擬大數加法, 關於這個char數組並不會自動初始化 假如 char s[35]; for(int i=0;i<10;++i) s[i]=i+‘0‘; 就會出現亂碼導致我一直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 大數+Trie