ACM-ICPC 2018 焦作赛区网络预赛 G. Give Candies

题解

题目大意 按照顺序分糖果 每个最少分1个 问有多少种情况

根据题意找到规律 答案为2的n-1次方 使用欧拉降幂和快速幂计算

AC代码

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 10;
const int MOD = 1e9 + 7;
char s[MAXN];

long long phi(long long x)
{
    long long ret = x;
    for (int i = 2; i * i <= x; ++i)
        if (x % i == 0)
        {
            ret -= ret / i;
            while (x % i == 0)
                x /= i;
        }
    if (x > 1)
        ret -= ret / x;
    return ret;
}
ll mpow(ll a, ll n, ll m) //a ^ n % m
{
    ll t = 1;
    while (n)
    {
        if (n & 1)
            t = (t * a) % m;
        a = (a * a) % m, n >>= 1;
    }
    return t;
}
int main()
{
#ifdef LOCAL
    //freopen("C:/input.txt", "r", stdin);
#endif
    int T;
    cin >> T;
    while (T--)
    {
        ll n = 0, p = phi(MOD);
        scanf("%s", s);
        for (int i = 0; s[i]; i++)
            n = (n * 10 + s[i] - '0') % p;
        n += p - 1;
        ll ans = mpow(2, n, MOD);
        cout << ans << endl;
    }

    return 0;
}