http://poj.org/problem?id=1995
2.Idea
Fast Power.
3.Source
ll mod_pow(ll x, ll n, ll mod)
{
if (n == 0) return 1;
ll res = mod_pow(x * x % mod, n / 2, mod);
if (n & 1) res = res * x % mod;
return res;
}
int main()
{
int z;
cin >> z;
while (z--) {
ll sum = 0;
ll mod;
cin >> mod;
int h;
cin >> h;
for (int i = 0; i < h; i++) {
ll a, b;
cin >> a >> b;
sum = (sum + mod_pow(a, b, mod)) % mod;
}
cout << sum << endl;
}
return 0;
}
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