HDU - 4704 Sum （费马小定理+快速幂）

Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 4328    Accepted Submission(s): 1759   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   |   We have carefully selected several similar problems for you:  6308 6307 6306 6305 6304  题目大意：输入一个数n，求pow（2，n-1）； 由于n非常大,所以这里要用到费马小定理:a^(p-1)%p == 1%p == 1;//p为素数 所以2^n%m == ( 2^(n%(m-1))*2^(n/(m-1)*(m-1)) )%m == (2^(n%(m-1)))%m * ((2^k)^(m-1))%m == (2^(n%(m-1)))%m;//k=n/(m-1) #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <cmath> using namespace std; typedef long long ll; const int mod = 1e9+7; char s[10000010]; /* 由于n非常大,所以这里要用到费马小定理:a^(p-1)%p == 1%p == 1;//p为素数 所以2^n%m == ( 2^(n%(m-1))*2^(n/(m-1)*(m-1)) )%m == (2^(n%(m-1)))%m * ((2^k)^(m-1))%m == (2^(n%(m-1)))%m;//k=n/(m-1) */ ll quick_mod(ll a,ll b){//快速幂 ll ans = 1; a %= mod; while(b){ if(b&1){ ans = ans*a%mod; b--; } b >>= 1; a = a*a %mod; } return ans; } int main() { while(cin >> s){//位数很大所以输入的是字符串 ll sum=0; int i; for(i=0;s[i];i++) sum = (sum*10%(mod-1)+s[i]-'0')%(mod-1); //将字符串处理成整数，注意最后得到的整数sum要除以mod-1,所以过程中要除余mod-1,而不是mod sum = (sum-1)%(mod-1); cout << quick_mod(2,sum) << endl; } return 0; }