UVa 10162 last Digit: Mathematical law

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem &problem=1103

1. First look at the rule of a single number:

0^n%10:0

1^n%10:1

2^n%10:2,4,8,6

3^n%10:3,9,7,1

4^n%10:4,6

5^n%10:5

6^n%10:6

7^n%10:7,9,3,1

8^n%10:8,4,2,6

9^n%10:9,1

2. Because the maximum cycle length is 4, we can see that the i^i%10 and i^ (i+20)%10 are the same, that is to say, the i^i%10 cycle is 20

and (1^1+......+20^20)%10=4,

So (1^1+......+40^40)%10=8,

(1^1+......+60^60)%10=2,

(1^1+......+80^80)%10=6,

(1^1+......+100^100)%10=0,

So it appears that S (N)%10=s (n%100)%10

So consider the latter two digits of N.

3. To simplify operations, further analysis is possible:

It may be calculated (1^1+......+10^10)%10=7, so (11^11+......+20^20)%10=7 (4-7+10=7 above), so that every 10 number and modulo 10 7,

Then we calculate the single digit of s (1) ~s (10) and the Single-digit a[i of [s (a)-s (]~[s)-S (10)],

Remember b=n%100, then ans= (b/10*7+a[b%20])%10

See more highlights of this column: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/

Complete code:

01./*0.015s*/
.  
#include <bits/stdc++.h>  
04.using namespace std;  
05.const int a[20] = {0, 1, 5, 2, 8, 3, 9, 2, 8, 7, 0, 1, 7, 0, 6, 1, 7, 4, 8, 7}; S (N)%10,0<=n<=19  
.  
07.char s[105];  
09.int Main ()  
10.{    int Len, b, C;    while (gets (s), s[0]!= ' 0 ')  
.    {        len = strlen (s);        if (len = = 1) b = s[0] &;        Else B = atoi (s + len-2);        printf ("%d\n", (B/10 * 7 + a[b%));  
.    return 0;  
20.}