#include<stdio.h>???//題目是：求的是a的b次方的最后一個(gè)數(shù)是什么
int?last(int?x,int?y){
x?=?x?%?10;
int?k;
int?a[5];?
switch(x){
case?1:
return?1;
case?5:
return?5;
case?6:
return?6;
case?0:
return?0;
case?2:
{
a[5]={2,4,8,6};
k?=?y?%?4;
return?a[k-1];
}
case?3:
{
a[5]={1,3,9,7};
k?=?y?%?4;
return?a[k];
}
case?4:
{
a[5]={4,6};
k?=?y?%?2;
return?a[k-1];
}
case?7:
{
a[5]={1,7,9,3};
k?=?y?%?4;
return?a[k];
}
case?8:
{
a[5]={8,4,2,6};
k?=?y?%?4;
return?a[k-1];
}
case?9:
{
a[3]={1,9};
k?=?y?%?2;
return?a[k];
}
}
}
int?main(){
int?a,b;
while(~scanf("%d?%d",&a,&b)){
if(b==0)
printf("1");
else
printf("%d",last(a,b));
}
return?0;
}
4 回答
已采納
慕移動(dòng)6052691
TA貢獻(xiàn)22條經(jīng)驗(yàn) 獲得超7個(gè)贊
修改后的程序：
#include <stdio.h>
int last(int x,int y){
x = x % 10;
int k;
//int a[5];?
switch(x){
case 1:
return 1;
break;
case 5:
return 5;
break;
case 6:
return 6;
break;
case 0:
return 0;
break;
case 2:
{
int a[4]={2,4,8,6};
k = y % 4;
return a[k-1];
};break;
case 3:
{
int a[4]={1,3,9,7};
k = y % 4;
return a[k];
};break;
case 4:
{
int a[2]={4,6};
k = y % 2;
return a[k-1];
};break;
case 7:
{
int a[4]={1,7,9,3};
k = y % 4;
return a[k];
};break;
case 8:
{
int a[4]={8,4,2,6};
k = y % 4;
return a[k-1];
};break;
case 9:
{
int a[2]={1,9};
k = y % 2;
return a[k];
};break;
}
}
int main(){
int a,b;
while(scanf("%d %d",&a,&b)){
if(b==0)
printf("1");
else
printf("%d",last(a,b));
}
return 0;
}
- 4 回答
- 0 關(guān)注
- 1473 瀏覽
添加回答
舉報(bào)
0/150
提交
取消