杭电--1061 Rightmost Digit
2017-04-11 19:26
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本题连接:点击打开链接
Total Submission(s): 55381 Accepted Submission(s): 20938
[align=left]Problem Description[/align]
Given a positive integer N, you should output the most right digit of N^N.
[align=left]Input[/align]
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).
[align=left]Output[/align]
For each test case, you should output the rightmost digit of N^N.
[align=left]Sample Input[/align]
2
3
4
[align=left]Sample Output[/align]
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.
[align=left]Author[/align]
Ignatius.L
[align=left]Recommend[/align]
We have carefully selected several similar problems for you: 1071 1425 1060 1205 1042
题意:先输入一个整数T,表示有T组测试数据,接下来给出一个数N(1<=N<=1,000,000,000),求该数的N次幂的个位是多少?
思路一:对于这一类涉及到大数,而要求的结果又比较简单,我在hdu--1005中就说过应该首先就要想到的是找规律,打个表观察一下,找出它的周期就好了,通常都是可以的。
代码一:
思路二:由于只要求个位数,所以有种方法很适用这个题,那就是快速幂取余(详解请点http://www.phpstudy.net/b.php/67443.html)。
快速幂算法明显依赖于以下公式:
快速幂取余代码如下:
本题代码二:
Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 55381 Accepted Submission(s): 20938
[align=left]Problem Description[/align]
Given a positive integer N, you should output the most right digit of N^N.
[align=left]Input[/align]
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).
[align=left]Output[/align]
For each test case, you should output the rightmost digit of N^N.
[align=left]Sample Input[/align]
2
3
4
[align=left]Sample Output[/align]
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.
[align=left]Author[/align]
Ignatius.L
[align=left]Recommend[/align]
We have carefully selected several similar problems for you: 1071 1425 1060 1205 1042
题意:先输入一个整数T,表示有T组测试数据,接下来给出一个数N(1<=N<=1,000,000,000),求该数的N次幂的个位是多少?
思路一:对于这一类涉及到大数,而要求的结果又比较简单,我在hdu--1005中就说过应该首先就要想到的是找规律,打个表观察一下,找出它的周期就好了,通常都是可以的。
代码一:
#include <iostream> #include <cstdio> using namespace std; /*void fun() //找出50以内的整数的N的N次方的个位数。 { for(int i=1; i<50; i++) { int a=i; for(int j=1; j<i; j++) //注意循环次数 { a=(a*i)%10; } printf("%d\t",a); } printf("\n************\n"); }*/ int main() { //fun(); //打表找规律 int t; scanf("%d",&t); while(t--) { int n,i,p; scanf("%d",&n); n=n%20; //周期为20 p=n; for(i=1; i<n; i++) p=(n*p)%10; printf("%d\n",p); } return 0; }
思路二:由于只要求个位数,所以有种方法很适用这个题,那就是快速幂取余(详解请点http://www.phpstudy.net/b.php/67443.html)。
快速幂算法明显依赖于以下公式:
快速幂取余代码如下:
int QY(int a,int b,int c) //求 a 的 b 次幂对 c 取的余数 { int k=1; //记录结果 a=a%c; //预处理一下,使a比c小 while(b) { if(b%2==1) //公式使用 k=(k*a)%c; b=b/2; a=(a*a)%c; } return k; }
本题代码二:
#include <cstdio> #include <cstring> #include <iostream> using namespace std; int qky(int a) { int k=1; int b=a; a=a%10; //本题对10取余,且为本身的次幂 while(b) { if(b%2==1) k=(k*a)%10; b/=2; a=(a*a)%10; } return k; } int main() { int t; scanf("%d",&t); while(t--) { int n; scanf("%d",&n); printf("%d\n",qky(n)); } return 0; }
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