D - Sum Time Limit:1000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u Submit Status Description Input 2 Output 2 Hint 1. For N = 2, S(1) = S(2) = 1. 2. The input file consists of multiple test cases. 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. 思路：首先根据规律找出sum=2^(n-1),转化为求2^(n-1)%mod利用费马小定理a^b%c=a^(b%(c-1))%c所以解决这个问题只需要两部1.用大数取余求得（N-1）%mod-12.用快速幂求得a^b%modac代码：#include #include #include #define mod 1000000007 using namespace std; char a[10000005]; long long quickmod(long long  a, long long  b) {     long long ans=1;     while(b)     {         if(b&1)         {             ans=(ans*a)%mod;         }         b=b/2;         a=(a*a)%mod;     }     return ans; } int main() {     long long sum,len;     while(gets(a))     {         len=strlen(a);         sum=0;         for(int i=0;i