Codeforces Round #260 (Div. 2)B. Fedya and Maths
2017-06-10 10:08
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B. Fedya and Maths
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Fedya studies in a gymnasium. Fedya's maths hometask is to calculate the following expression:
(1n + 2n + 3n + 4n) mod 5
for given value of n. Fedya managed to complete the task. Can you? Note that given number n can
be extremely large (e.g. it can exceed any integer type of your programming language).
Input
The single line contains a single integer n (0 ≤ n ≤ 10105).
The number doesn't contain any leading zeroes.
Output
Print the value of the expression without leading zeros.
Sample test(s)
input
output
input
output
Note
Operation x mod y means taking remainder after division x by y.
Note to the first sample:
让求出 (1^n + 2^n + 3^n + 4^n )mod 5 的结果,而n最大能够有10^5位数。仅仅能用字符串来存,将取模分开得到
1^n mod 5 = 1 ;
2^n mod 5 = 随着n从0開始,结果为 1 2 4 3 1 2 4 3 1 2 4 3 。。
。。以1 2 4 3 循环
3^n mod 5 = 随着n从0開始。结果为 1 3 4 2 1 3 4 2 1 3 4 2.。
。
。。以1 3 4 2 循环
4^n mod 5 = 。。
。。结果为 1 4 1 4 1 4 以1 4 循环
求给出的n是奇数还是偶数,方法推断最后一位奇数偶数。
再求解n mod 4 。方法求出最后两位的对4取模的值
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Fedya studies in a gymnasium. Fedya's maths hometask is to calculate the following expression:
(1n + 2n + 3n + 4n) mod 5
for given value of n. Fedya managed to complete the task. Can you? Note that given number n can
be extremely large (e.g. it can exceed any integer type of your programming language).
Input
The single line contains a single integer n (0 ≤ n ≤ 10105).
The number doesn't contain any leading zeroes.
Output
Print the value of the expression without leading zeros.
Sample test(s)
input
4
output
4
input
124356983594583453458888889
output
0
Note
Operation x mod y means taking remainder after division x by y.
Note to the first sample:
让求出 (1^n + 2^n + 3^n + 4^n )mod 5 的结果,而n最大能够有10^5位数。仅仅能用字符串来存,将取模分开得到
1^n mod 5 = 1 ;
2^n mod 5 = 随着n从0開始,结果为 1 2 4 3 1 2 4 3 1 2 4 3 。。
。。以1 2 4 3 循环
3^n mod 5 = 随着n从0開始。结果为 1 3 4 2 1 3 4 2 1 3 4 2.。
。
。。以1 3 4 2 循环
4^n mod 5 = 。。
。。结果为 1 4 1 4 1 4 以1 4 循环
求给出的n是奇数还是偶数,方法推断最后一位奇数偶数。
再求解n mod 4 。方法求出最后两位的对4取模的值
#include <cstdio> #include <cstring> char str[110000] ; int main() { int ans = 1 , l , i , k ; scanf("%s", str); l = strlen(str); if( (str[l-1]-'0')%2 ) ans += 4 ; else ans += 1 ; if( l-1 != -1 ) k = ( str[l-1]-'0' ); if( l-2 != -1 ) k = k + (str[l-2]-'0')*10 ; if( k%4 == 0 ) ans += 2 ; else if( k%4==1 ) ans += 5 ; else if( k%4 ==2 ) ans += 8 ; else ans += 5 ; printf("%d\n", ans%5); return 0; }
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