CSU 1347 Last Digit-优快云博客

本文链接：https://blog.youkuaiyun.com/gtuif/article/details/79933930

本文介绍了一种高效算法，用于计算特定形式的数学函数f(n,k) = 1^k + 2^k + ... + n^k的最后一个数字。通过预先计算并利用模幂运算，该算法能在大规模输入范围内快速得出结果。

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The function f(n, k) is defined by f(n, k) = 1^k + 2^k + 3^k +...+ n^k. If you know the value of n and k, could you tell us the last digit of f(n, k)?
For example, if n is 3 and k is 2, f(n, k) = f(3, 2) = 1² + 2² + 3² = 14. So the last digit of f(n, k) is 4.

Input

The first line has an integer T (1 <= T <= 100), means there are T test cases.
For each test case, there is only one line with two integers n, k (1 <= n, k <= 10⁹), which have the same meaning as above.

Output

For each test case, print the last digit of f(n, k) in one line.

Sample Input

10
1 1
8 4
2 5
3 2
5 2
8 3
2 4
7 999999997
999999998 2
1000000000 1000000000

Sample Output

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
ll mod_pow(ll x,ll n,ll mod){
	ll res=1;
	while(n>0){
		if(n&1) res=res*x%mod;
		x=x*x%mod;
		n>>=1;
	}
	return res;
}
int main()
{
	int T;
	scanf("%d",&T);
	ll a[10];
	//printf("==%lld\n",mod_pow(0,5,10));
	while(T--){
		ll n,k;
		scanf("%lld%lld",&n,&k);
		//memset(a,0,sizeof(a));
		ll sum=0;
		for(int i=0;i<10;i++){
			a[i]=mod_pow(i,k,10);
			//printf("++%lld\n",a[i]);
			sum=(sum+a[i])%10;
		}
		//printf("---%lld\n",sum);
		ll res=0;
		res=n/10*sum%10;
		//printf("res=%lld\n",res);
		for(int i=0;i<=n%10;i++){
			res+=a[i];
		}
		printf("%lld\n",res%10);
	}
	
	return 0;
}