错哪了

[木可大大]

Given a positive integer N, you should output the most right digit of N^N.

Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

Output
For each test case, you should output the rightmost digit of N^N.

Sample Input
2
3
4

Sample Output
7
6

Hint
In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7.
In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.
我的代码是
#include<stdio.h>
#include<stdlib.h>
int main()
{
unsigned n,m,temp,flag;
while(1)
{
scanf("%d",&n);
while(m>0)
{
scanf("%d\n",&m);
temp=m^m;
while(temp/10>0)
{
flag=temp%10;
}

print("%d\n",flag);
}
}
return 0;
}错哪里了

[rollrock1987]

can you。。。chinese？？？

[青葱岁月不可追]

看见英语就头疼

[bshawk]

你代码主要错误在：m^m并不是m的m次方预算，而是m和m做异或运算！

如果你只是简单的验证下：可以用如下代码

#include <stdio.h> 
#include <stdlib.h> 
#include <math.h> 
int main() 
{ 
	unsigned m,temp,flag; 
	while(1) 
	{ 
		scanf("%d",&m); 
		if(m>0) 
		{
			temp=pow(m,m); 
			flag =temp%10;
			printf("%d\n",flag); 
		}
	} 
	return 0; 
}

但是上面的代码并不是本题要考你的地方，本题要考你是对很大的数（1,000,000,000）你该如何来实现？ 显然，上面的算法对大数是不行的，需要重新考虑算法！ 这怎么有点像ACM上的练习题？

Good Luck！

[expert2000]

#include <stdio.h>
#include <stdlib.h>
int main()
{
unsigned int n,m,temp,flag;
n = 0;
m = 0;
temp = 1;
//while(1)
//{
// scanf("%d",&n);
/// while(m>0)
//{
scanf("%d",&m);
while(n<m)
{
temp=temp*m;
n++;
}
printf("%d\n",temp);
if(temp/10>0)
{
flag=temp%10;
}

printf("%d\n",flag);
//}
//}
return 0;
}

[星羽]

#include "stdlib.h"

int get_last(int d)
{
    while ((d %= 10) >= 10)
        ;
    return d;
}

int main()
{
    int* data = 0;
    int  num = 0;
    int  i = 0;

    printf("please input the data : \n");
    scanf("%d", &num);

    if (num <= 0)
        return 1;

    data = (int*)malloc(sizeof(int) * num);
    
    while (i < num)
        scanf("%d", &data[i++]);

    printf("the result is : \n");

    for (i = 0; i < num; ++i)
    {
        int d = data[i];
        int j = 0;
        int w = get_last(d);
        int r = w;

        while (++j < d)
            r = get_last(r * w);
        
        printf("%d\n", r);
    }

    free(data);

    return 0;
}


-------------

please input the data :
5
1
2
3
4
12345678
the result is :
1
4
7
6
4
请按任意键继续. . .

[zpnjit]

赞成楼上的

[rollrock1987]

print("%d\n",flag); 你这个 是什么函数？？

不是 printf吗？？

[ustczwx]

错的太多了，第一行输入是测试数字的个数，后面包含N行输入。
unsiged int n,m,temp,result;
scanf("%d\n", &n);
while (n>0)
{
scanf("%d\n", &m);
result=temp=m%10;
if (temp==0||temp==1||temp==5||temp==6)
{printf("%d\n",temp); /*all these digits will be same as input*/}
else
{
result = 1;
while(m>0)
{
if (m & 0x01 == 0x01) /* m mod by 2 */
result = (result * temp) % 10;
temp = (temp * temp) % 10;
m = m >> 1; /* m divided by 2 */
}
printf("%d\n",result);
}
n--; /*count decrease*/
}

[kkndciapp]

void main()
{
int n,num;
cout<<"input n:"<<endl;
cin>>n;
for(int i=0;i<n;i++)
{
cin>>num;
int temp=num%10;
for(int j=0;j<num-1;j++)
{

temp*=num;
temp=temp%10;
}
cout<<temp<<endl;;
}

}

[迷途的书童]

感觉4楼分析的很不错啊！俺把它改了一下；
如n^n;
先求m＝n％10的最后一位；
再m=m×n%10求最后一位；
。。。。一共计算n次；代码如下：

#include "stdafx.h"
#include <stdio.h> 
#include <stdlib.h> 
#include <math.h> 


int main(int argc, char* argv[])
{
	 int m[100],temp,nNum; 
    int i=0; 
	printf("input the sample:\n");
    scanf("%d",&nNum);
    while(nNum--) 
    { 
        scanf("%d",&m[i++]); 
        
    } 
	printf("output thr right most digit :\n");
	for(int j=0;j<i;j++)
	{    temp=1;
		for(int k=0;k<m[j];k++)
		{
			temp*=m[j];
			temp=temp%10;
		}
		printf("%d\n",temp);


}
	return 0;

}

[ustczwx]

楼主一定要试一下我9楼的算法，效率可不是一般的高啊，嘿嘿

[fanzf24]

好像是浙大ACM的题目，注释是修改的部分

#include <stdio.h>
#include <stdlib.h>
int main()
{
unsigned n,m,temp,flag;
while(1)
{
scanf("%d",&n); /*scanf("%d",&m);*/
while(m>0)
{
scanf("%d\n",&m);
temp=m^m;
while(temp/10>0)
{
flag=temp%10;
}

print("%d\n",flag); /*printf("%d/n",flag);*/
}
}
return 0;
}

[mifeixq]

写了个效率高点的…… 你看看能AC不…… 不能我再改改

#include <stdio.h> 
#include <math.h>
int get_result(int m, int n)
{
	int store;
	int count;
	if(m==0||m==1||m==5||m==6)
		return m;
	switch(m)
	{
		case 2:
			store = n%4;
			if(n>4)
				return pow(2, store)*6;
			else
				return pow(2, store);
		case 3:
			store = n%4;
			return pow(3, store);
		case 4:
			store = n%2;
			if(store)
				return 4;
			else
				return 6;
		case 7:
			store = n%4;
			return pow(7, store);
		case 8:
			store = n % 4;
			return pow(8, store)*6;
		case 9:
			store = n%2;
			if(store)
				return 9;
			else
				return 1;
	}
}
int main()
{
	int count,num;
	scanf("%d",&count);
	while(count--)
	{
		int result;
		int last;
		scanf("%d",&num);
		last = num % 10;
		result=get_result(last, num)%10;
		printf("%d\n", result);
	}

	return 0;
}