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  • HDU-4099Revenge of Fibonacci (高精度加法&&字典树)

    传送门

    11年上海区域赛题

    T组数据,求对于每个数,求最小的n使得该数为斐波那契数列第n位的前缀,若n大于100000,输出-1

     1 #include <cstdio>
     2 #include <cstring>
     3 #include <algorithm>
     4 using namespace std;
     5 const int maxn = 4e6 + 10;
     6 const int maxsize = 10;
     7 
     8 int ch[maxn][maxsize];
     9 int val[maxn];
    10 int A[60];
    11 int B[60];
    12 char s[100];
    13 
    14 int T;
    15 
    16 struct Trie {
    17     int sz;
    18     Trie() {
    19         sz = 1;
    20         memset(ch[0], 0, sizeof(ch[0]));
    21         memset(val, -1, sizeof(val));
    22     }
    23     int idx(char c) {return c - '0';}
    24 
    25     void insert(int* a, int len, int v) {
    26         int u = 0;
    27         for (int i = len - 1; i >= max(0, len - 40); i--) {
    28             int c = a[i];
    29             if (!ch[u][c]) {
    30                 memset(ch[sz], 0, sizeof(ch[sz]));
    31                 ch[u][c] = sz++;
    32             }
    33             u = ch[u][c];
    34             if (val[u] == -1) {
    35                 val[u] = v;
    36             }
    37         }
    38     }
    39 
    40     int find(char* s) {
    41         int len = strlen(s);
    42         int u = 0;
    43         for (int i = 0; i < len; i++) {
    44             int c = idx(s[i]);
    45             if (!ch[u][c]) {
    46                 return -1;
    47             }
    48             u = ch[u][c];
    49         }
    50         return val[u];
    51     }    
    52 
    53 };
    54 
    55 Trie Tree;
    56 
    57 
    58 int main() {
    59     memset(A, 0, sizeof(A));
    60     memset(B, 0, sizeof(B));
    61     B[0] = 1;
    62     int len1 = 1;
    63     int len2 = 1;
    64     for (int n = 0; n < 100000; n++) {
    65         Tree.insert(B, len2, n);
    66         int len3 = len2;
    67         for (int i = 0; i < len3; i++) {
    68             A[i] += B[i];
    69             if (A[i] >= 10) {
    70                 A[i] -= 10;
    71                 A[i + 1]++;
    72             }
    73         }
    74         if (A[len3]) len3++;
    75         if (len3 == 51) {
    76             len3--;
    77             for (int i = 0; i < len3; i++) {
    78                 A[i] = A[i + 1];
    79             }
    80             A[50] = 0;
    81             len2--;
    82             for (int i = 0; i < len2; i++) {
    83                 B[i] = B[i + 1];
    84             }
    85             B[len2] = 0;
    86         }
    87 
    88         swap(A, B);
    89         len1 = len2;
    90         len2 = len3;
    91 
    92     }
    93     scanf("%d", &T);
    94     for (int i = 1; i <= T; i++) {
    95         scanf("%s", s);
    96         printf("Case #%d: %d
    ", i, Tree.find(s));
    97     }
    98     return 0;
    99 }
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  • 原文地址:https://www.cnblogs.com/xFANx/p/8521619.html
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