洛谷P1630求和

刷题之路-数论

洛谷P1630求和

传送门

做题思路

核心：(a + b) ^ x ≡ a ^ x (% b)
因此 1 ^ b ≡ (1 + mod) ^ b (% mod)

代码

#include<iostream>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<functional>
#include<cmath>
#include<cstdlib>
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod = 10000;
typedef pair<int, int> pii;
typedef long long ll;
const int N = 50005;
const int M = 1000005;
int ans[N];
ll qmi(ll a, int b)
{
    
    
	ll ans = 1 % mod;
	while (b)
	{
    
    
		if (b & 1) ans = ans * a % mod;
		a = a * a % mod;
		b >>= 1;
	}
	return ans;
}
int main()
{
    
    
	int t;
	cin >> t;
	while (t--)
	{
    
    
		int a, b;
		cin >> a >> b;
		ll ans = 0;
		for (int i = 1; i <= min(a, mod); i++) ans = (ans + qmi(i, b)) % mod;
		if (a > mod)
		{
    
    
			int num = a / mod;
			ans = ans * num % mod;
			for (int i = 1; i <= a % mod; i++) ans = (ans + qmi(i, b)) % mod;
		}
		cout << ans << endl;
	}
	return 0;
}