ZOJ 3903 Ant(公式推导)

这个公式推导过程是看的这位大牛的http://blog.csdn.net/bigbigship/article/details/49123643

 扩展欧几里德求模的逆元方法：

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

using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    ll r = exgcd(b, a % b, x, y);
    ll t = x % mod;
    x = y % mod;
    y = ((t - a / b * y) % mod + mod) % mod;
    return r;
}
int main()
{
    ll x, y;
    ll inv2, inv4, inv6;
    exgcd(2, mod, inv2, y);
    exgcd(4, mod, inv4, y);
    exgcd(6, mod, inv6, y);
    ll T, n;
    printf("%lld, %lld, %lld\n", inv2, inv4, inv6);
    scanf("%lld", &T);
    while (T--)
    {
        scanf("%lld", &n);
        n %= mod;
        ll ans = (n * n % mod + n) % mod * n % mod * n % mod * inv2 % mod;
        ans += ((n * (n + 1) % mod) * n % mod * (n + 1) % mod * inv2 % mod) % mod;
        ans += (n + 2) * n % mod * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod;
        ans =(((ans - n * n % mod * (n + 1) % mod * (n + 1) % mod * inv4 % mod + mod) % mod + mod) % mod);
        printf("%lld\n", ans);
    }
    return 0;
}

费马小定理求模的逆元法

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

using namespace std;
typedef long long ll;
const ll mod = 1e9 + 7;
ll power_mod(ll a, ll b, ll mod)
{
    ll ans = 1;
    while (b)
    {
        if (b & 1) ans = ans * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}
int main()
{
    ll inv2 = power_mod(2, mod - 2, mod);
    ll inv4 = power_mod(4, mod - 2, mod);
    ll inv6 = power_mod(6, mod - 2, mod);
    ll T, n;
    scanf("%lld", &T);
    while (T--)
    {
        scanf("%lld", &n);
        n %= mod;
        /*ll ans = (n * n % mod + n) % mod * n % mod * n % mod * inv2 % mod;
        ans += ((n * (n + 1) % mod) * n % mod * (n + 1) % mod * inv2 % mod) % mod;
        ans += (n + 2) * n % mod * (n + 1) % mod * (2 * n + 1) % mod * inv6 % mod;
        ans =(((ans - n * n % mod * (n + 1) % mod * (n + 1) % mod * inv4 % mod + mod) % mod + mod) % mod);*/
        ll ans = n * (n + 1) % mod * n % mod * n % mod * inv2 % mod;
        ans = (ans + n * (n + 1) % mod * n % mod * (n + 1) % mod * inv2 % mod) % mod;
        ans = (ans + n * (n + 1) % mod * (n + 2) % mod * (2 * n + 1) % mod * inv6 % mod) % mod;
        ans = (ans - n * n % mod * (n + 1) % mod * (n + 1) % mod * inv4 % mod + mod) % mod;
        printf("%lld\n", ans);
    }
    return 0;
}