C - Last Digit

Description

    The functionf(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 off(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 off(n, k) is 4.

Input

    The first line has an integerT (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 off(n, k) in one line.

Sample Input

101 18 42 53 25 28 32 47 999999997999999998 21000000000 1000000000

Sample Output

1234567890
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]);}}