HDU-1061-Rightmost Digit
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Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33177 Accepted Submission(s): 12701
Problem Description
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).
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
234
Sample Output
76HintIn 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.以前做的,之前没做出来,直接用的循环,TLE了,应该是常理之中吧,然后现在想想这种题就是要去找规律的,突然发现自己进步了不少,这题放在以前都不知道会从这种方面想。要输出最右边一位,并且是N^N的最右一位,先举出一下例子,如11, 以1结尾,乘多少次都是1, 同样的还有0, 5 , 6。其他的也有规律,他们乘以好多次都会回到原来的数,如7 : 7 -〉 9 -〉 3 -〉 1 -〉 7 就是这个规律了。这题还是很水的,贴个代码:#include <cstdio>#include <cstring>using namespace std;int main(){ int T, s, N; scanf("%d",&T); while(T--) { scanf("%d",&N); s=N%10; if(s == 0 || s == 1 || s == 5 || s == 6) { printf("%d\n", s); } else if(s == 2) { if(N%4==1)printf("2\n"); if(N%4==2)printf("4\n"); if(N%4==3)printf("8\n"); if(N%4==0)printf("6\n"); } else if(s == 3) { if(N%4==1)printf("3\n"); if(N%4==2)printf("9\n"); if(N%4==3)printf("7\n"); if(N%4==0)printf("1\n"); } else if(s == 4) { if(N%2==1)printf("4\n");if(N%2==0)printf("6\n"); } else if(s == 7) { if(N%4==1)printf("7\n");if(N%4==2)printf("9\n"); if(N%4==3)printf("3\n");if(N%4==0)printf("1\n"); } else if(s == 8) { if(N%4==1)printf("8\n");if(N%4==2)printf("4\n"); if(N%4==3)printf("2\n");if(N%4==0)printf("6\n"); } else if(s == 9) { if(N%2==1)printf("9\n"); if(N%2==0)printf("1\n"); } } return 0;}
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