C - Last Digit

最新推荐文章于 2021-09-20 12:36:34 发布

原创最新推荐文章于 2021-09-20 12:36:34 发布 · 606 阅读

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本文介绍如何通过编程计算给定整数n和k时，函数f(n,k) = 1^k + 2^k + ... + n^k 的最后一位数字。通过观察循环周期，我们能够简化计算过程，利用预定义数组存储结果来提高效率。

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Description

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

1
2
3
4
5
6
7
8
9
0
打表发现n的循环周期可以是100，所以直接n%100,发现m的循环周期可以是4，所以只要在a[100]储存4个数
#include<stdio.h>
#include<string.h>
int a[100];
int quick(int a,int b)
{
	int ans=1;
	a=a%10;
    while(b>0)
    {
    	if(b%2==1)
    	ans=ans*a%10;
    	b=b/2;
    	a=(a*a)%10;
    }
    return ans;
}

int main()
{
	int i,n,k,t,j,T;
	scanf("%d",&T);
	while(T--)
	{
	 scanf("%d%d",&n,&k);
	 n=n%100;
	 for(i=1;i<=4;i++)
	 {
	   t=0;
	   for(j=1;j<=n;j++)
	   {
		t=(t+quick(j,k))%10;
	   }
	   a[i]=t;
	 }
	 a[0]=a[4];
	 k=k%4;
     printf("%d\n",a[k]);
	}
	return 0;
}

另一种是先找循环节，再算
#include<stdio.h>
#include<string.h>
int powermod(int n,int k){
	int ans=1;
	n=n%10;
	while(k){
		if(k%2) ans=(ans*n)%10;
		k=k/2;
		n=(n*n)%10;
	}
	return ans;
}
int main(){
	int T,i,j,n,k;
	scanf("%d",&T);
	while(T--){
		int f[1111]={0};
		int ans=0,t=0;
		scanf("%d%d",&n,&k);
		for(i=1;i<=1000;i++){
			t=powermod(i,k);
			f[i]=(t+f[i-1])%10;
		}
		int temp,flag;
		for(i=1;i<=1000;i++){
			flag=1;
			for(j=i+1;j<=1000;j++){
				if(f[j]!=f[j%i]) {flag=0; break;}
			}
			if(flag) {temp=i; break;}
		}
		ans=n%temp;
		printf("%d\n",f[ans]);
	}
}