Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 5322 Accepted Submission(s): 2099

Problem Description

Sample Input

2

Sample Output

2
Hint

For N 2, S(1) S(2) 1.

The input file consists of multiple test cases.

Source
2013 Multi-University Training Contest 10

费马小定理/快速幂运算
https://blog.csdn.net/codeswarrior/article/details/79867198

#include bits/stdc .h using namespace std;const long long mod 1e9 7; char r[100010];long long qpow(long long x,long long k) long long ans 1,res x; while(k) if(k 1) ans (ans*res)%mod; k 1; res res*res%mod; return ans;int main(){ long long n; while(~scanf( %s ,r)) { n 0; for(int i 0;i strlen(r);i ) n (n*10 r[i]- 0 )%(mod-1); long long ans qpow(2,n-1); printf( %lld\\n ,ans);