hdu 4704 Sum（数论）

题目链接：hdu 4704 Sum

解题思路

费马小定理，指数的循环节为mod-1

代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 5;

struct bign {
    int len, num[maxn];

    void delzero() {
        while (len && num[len-1] == 0) len--;
        if (len == 0) num[len++] = 0;
    }
    void operator = (char* number) {
        len = strlen(number);
        for (int i = 0; i < len; i++)
            num[i] = number[len-i-1] - '0';
        delzero();
    }
    void operator -- () {
        if (num[0] == 0 && len == 1) return;
        int s = -1;
        for (int i = 0; i < len; i++) {
            s = s + num[i];
            num[i] = (s + 10) % 10;
            if (s >= 0) break;
        }
        delzero();
    }
    int operator % (const int& k) {
        int s = 0;
        for (int i = len-1; i >= 0; i--)
            s = (10LL * s + num[i]) % k;
        return s;
    }
};

char str[maxn];
bign N;

int pow_mod(int x, int n) {
    int ret = 1;
    while(n) {
        if (n&1) ret = 1LL * ret * x % mod;
        x = 1LL * x * x % mod;
        n >>= 1;
    }
    return ret;
}

int main () {
    while (scanf("%s", str) == 1) {
        N = str;
        --N;
        int k = N % (mod-1);
        printf("%d\n", pow_mod(2, k));
    }
    return 0;
}