HDU 4704 Sum （2013多校10,1009题）

 

Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 94    Accepted Submission(s): 41

Problem Description

 

 

Sample Input

2

 

 

Sample Output

2

Hint

1. For N = 2, S(1) = S(2) = 1. 2. The input file consists of multiple test cases.

 

 

Source

2013 Multi-University Training Contest 10

 

 

Recommend

zhuyuanchen520

 

 

 

 

其实这题就是求2^(n-1) % MOD;

可先对(n-1)%(MOD-1)

 

这样就很容易求了

 

 1 /* ***********************************************

 2 Author        :kuangbin

 3 Created Time  :2013/8/22 12:00:50

 4 File Name     :F:\2013ACM练习\2013多校10\1009.cpp

 5 ************************************************ */

 6 

 7 #include <stdio.h>

 8 #include <string.h>

 9 #include <iostream>

10 #include <algorithm>

11 #include <vector>

12 #include <queue>

13 #include <set>

14 #include <map>

15 #include <string>

16 #include <math.h>

17 #include <stdlib.h>

18 #include <time.h>

19 using namespace std;

20 

21 const int MOD = 1e9+7;

22 long long n;

23 long long solve()

24 {

25     long long ret = 1;

26     long long tmp = 2;

27     while( n > 0)

28     {

29         if(n%2)

30         {

31             ret *= tmp;

32             ret %= MOD;

33         }

34         tmp *= tmp;

35         tmp %= MOD;

36         n = n/2;

37     }

38     return ret;

39 }

40 char str[5000000];

41 int main()

42 {

43     //freopen("in.txt","r",stdin);

44     //freopen("out.txt","w",stdout);

45     while(scanf("%s",str) == 1)

46     {

47         n = 0;

48         int len = strlen(str);

49         for(int i = 0;i < len;i++)

50         {

51             n = (n*10 + str[i] -'0')%(MOD-1);

52         }

53         n = (n-1+MOD-1)%(MOD-1);

54         printf("%d\n",(int)solve());

55     }

56     return 0;

57 }