HDU 4704 SUM(费马小定理)

题意：求2的n-1次方

解题思路：费马小定理

解题代码：

 1 // File Name: 4704.c
 2 // Author: darkdream
 3 // Created Time: 2013年09月08日 星期日 20时32分42秒
 4 
 5 #include<stdio.h>
 6 #include<string.h>
 7 #include<stdlib.h>
 8 #include<time.h>
 9 #include<math.h>
10 #define LL long long
11 #define maxn 100009
12 #define mod 1000000007
13 //freopen("/home/plac/problem/input.txt","r",stdin);
14 //freopen("/home/plac/problem/output.txt","w",stdout);
15 
16 
17 char str[maxn];
18 LL a[maxn];
19 LL POW (LL n ,LL k )
20 {
21    LL ans = 1;
22    LL p = n;
23    while(k)
24    {
25        if(k &1)
26            ans =(ans * p)%mod ;
27        p = p*p %mod;
28        k >>=1;
29    }
30    return ans;
31 }
32 LL len ;
33 LL chu(LL k){
34    LL temp = 0;
35    for(int i = len;i >=1 ;i --)
36    {
37       temp = temp*10 + a[i];
38       if(temp > k )
39         temp = temp%k;
40    }
41  return temp;
42 }
43 void jian()
44 {
45   for(int i = 1;i <= len ;i ++)
46   {
47      if(a[i] >= 1)
48      {
49       a[i] -= 1;
50        return;
51      }
52      else 
53          a[i] = 9;
54   }
55 }
56 int main(){
57     while(scanf("%s",&str[1]) != EOF)
58     {
59        len = strlen(&str[1]);
60       for(int i = 1; i<= len ;i ++)
61           a[i] = str[len-i+1] - '0';
62        jian();    
63        LL k = chu(mod-1);
64        printf("%lld\n",POW(2,k));
65     }
66 
67 return 0 ;
68 }

View Code

posted on 2013-09-08 20:58 dark_dream 阅读(232) 评论(0) 编辑收藏举报