Approximations
| Time Limit: 2000MS | Memory Limit: 131072K | |
| Total Submissions: 419 | Accepted: 23 |
Description
For any decimal fraction, we can obtain a set of approximations of different accuracy by mean of rounding. Take 0.2503 for example, we have the following approximations:
- 0.2503
- 0.250
- 0.25
- 0.3
- 0.
If two fractions A and B can both be rounded to C, we call C a common approximation of A and B. Two fractions may have more than one common approximations, each having a distinct accuracy. For example, 0.2503 and 0.2504 have common approximations 0.250 and 0.25. The accuracy of the former is 10?3, while that of the latter is 10?2. Among all common approximations of two fractions, there is one that has the highest accuracy, and we call it the most accurate common approximation (MACA) of the two fractions. By this definition, the MACA of 0.2503 and 0.2504 is 0.250.
Given N fractions Ai (1 ≤ i ≤ N) in the range [0, 0.5), find a fraction x that maximizes the sum of ?log10 (the accuracy of the MACA of Ai and x). Report that maximized sum.
Input
The first line contains one integer N. N ≤ 100000.
Each of the next N lines contains a decimal fraction Ai. The total number of digits of the N decimal fractions doesn‘t exceed 400000. There is always a radix point, so zero is "0." instead of "0".
Output
One integer, the maximized sum.
Sample Input
4 0.250 0.2506 0.25115 0.2597
Sample Output
11
Hint
Source
董华星在他09年的论文里说可以用字典树写。。我试着写了下，然而感觉题目是不是给的范围有问题啊。测了好多样例没问题。欢迎各位大佬给个样例测测0 0。
代码如下：
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #define clr(x) memset(x,0,sizeof(x)) 5 #define clr_1(x) memset(x,-1,sizeof(x)) 6 #define LL long long 7 #define mod 1000000007 8 #define INF 0x3f3f3f3f 9 #define next nexted 10 using namespace std; 11 const int N=1e5+10; 12 const int M=4e5+10; 13 int next[M][10]; 14 int num[M]; 15 char s[M]; 16 int n,m,k; 17 int root,ttot; 18 void tadd(char *s,int root) 19 { 20 int now=root,p; 21 int len=strlen(s); 22 for(int i=0;i<len;i++) 23 { 24 p=s[i]-‘0‘; 25 if(next[now][p]==0) 26 { 27 next[now][p]=++ttot; 28 } 29 now=next[now][p]; 30 num[now]++; 31 } 32 return ; 33 } 34 int ans,prenode; 35 void dfs(int now,int prenum) 36 { 37 int nownum,p; 38 for(int i=0;i<10;i++) 39 { 40 nownum=prenum; 41 p=next[now][i]; 42 if(i==0) 43 { 44 if(prenode!=0) 45 { 46 prenode=next[prenode][9]; 47 if(prenode!=0) 48 { 49 for(int j=5;j<10;j++) 50 if(next[prenode][j]!=0) 51 nownum+=num[next[prenode][j]]; 52 } 53 } 54 } 55 else 56 { 57 if(next[now][i-1]!=0) 58 for(int j=5;j<10;j++) 59 { 60 if(next[next[now][i-1]][j]!=0) 61 { 62 nownum+=num[next[next[now][i-1]][j]]; 63 if(i-1>=5) 64 { 65 nownum+=num[next[next[now][i-1]][j]]; 66 } 67 } 68 69 } 70 } 71 if(i!=0) 72 { 73 prenode=next[now][i-1]; 74 } 75 if(p!=0) 76 { 77 nownum+=num[p]; 78 for(int j=5;j<10;j++) 79 { 80 if(next[p][j]!=0) 81 { 82 nownum-=num[next[p][j]]; 83 } 84 } 85 if(i>=5) 86 nownum+=num[p]; 87 if(nownum>ans) 88 ans=nownum; 89 // cout<<i<<" "<<nownum<<endl; 90 dfs(p,nownum); 91 } 92 else 93 { 94 if(nownum>ans) 95 ans=nownum; 96 } 97 } 98 return ; 99 } 100 int main() 101 { 102 scanf("%d",&n); 103 ttot=root=0; 104 for(int i=1;i<=n;i++) 105 { 106 scanf("%s",s); 107 tadd(s+2,root); 108 } 109 ans=0; 110 dfs(root,0); 111 printf("%d\n",ans); 112 return 0; 113 }
