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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; }
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原文地址：http://www.cnblogs.com/Howe-Young/p/4932565.html
