Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1465 Accepted Submission(s): 622

Problem Description

Sample Input

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

#include 
  
   
#include 
   
     #include 
    
      #include 
     
       #include 
      
        using namespace std; typedef long long int LL; const LL mod=1e9+7; const LL phi=1e9+6; char str[200000]; LL quickpow(LL x) { if(x==0LL) return 1LL; LL e=2,ret=1; while(x) { if(x%2) ret=(ret*e)%mod; e=(e*e)%mod; x/=2LL; } return ret; } int main() { while(scanf("%s",str)!=EOF) { int len=strlen(str); /// N - 1 if(str[len-1]>'0') { str[len-1]--; } else { int p=len-1; while(str[p]=='0') { str[p]='9'; p--; } str[p]--; } LL ret=0; for(int i=0;i
       
        =phi) ret-=phi; } cout<

HDOJ 4704 Sum 规律 欧拉定理

Sum

HDOJ 4704 Sum 规律欧拉定理