2146. 【2017.6.17普及组模拟】小明解密码
题目大意:
求t个 n^m的个位数。
方法一(找规律):
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#define fre(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
using namespace std;
const int MAX=2147483647;
int f[20][20];
int main()
{
//fre(a);
f[2][1]=2,f[2][2]=4,f[2][3]=8,f[2][0]=6;
f[3][1]=3,f[3][2]=9,f[3][3]=7,f[3][0]=1;
f[7][1]=7,f[7][2]=9,f[7][3]=3,f[7][0]=1;
f[8][1]=8,f[8][2]=4,f[8][3]=2,f[8][0]=6;
f[4][1]=4,f[4][0]=6;f[9][1]=9,f[9][0]=1;
int t;
scanf("%d",&t);
while(t--)
{
long long n,m;
scanf("%lld%lld",&n,&m);
n%=10;
if(n==0||n==1||n==5||n==6) printf("%d\n",n);
else if(n==2||n==3||n==7||n==8) printf("%d\n",f[n][m%4]);
else printf("%d\n",f[n][m%2]);
}
return 0;
}
方法二(快速幂):
将一个幂的指数分成两份如:n^10 =n^5 ×n^5
(因为都是10个n相乘),但当指数为奇数时,就要多成一个n,如:n^11 =n^5 ×n^5 ×n
#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#define fre(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
using namespace std;
const int MAX=2147483647;
long long n,m,ans,t;
void ksm(long long x)
{
if(x==1)
{
ans=n;
return;
}
ksm(x/2);
int d=1;
if(x&1) d=n;
ans=(ans*ans*d)%10;
}
int main()
{
//fre(a);
scanf("%lld",&t);
while(t--)
{
scanf("%lld%lld",&n,&m);
n%=10;
ksm(m);
printf("%lld\n",ans);
}
return 0;
}