codeforces #260 B. Fedya and Maths(水)
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B. Fedya and Maths
点击打开题目链接
time limit per test
1 secondmemory limit per test
256 megabytesinput
standard inputoutput
standard outputFedya studies in a gymnasium. Fedya's maths hometask is to calculate the following expression:
for given value of n. Fedya managed to complete the task. Can you? Note that given numbern 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 divisionx byy.
Note to the first sample:
水题,只要判断输入的n是否能被4整除即可(如果一个数可以被4整除,则只要这个数的最后两位被4整除即可,证明,请看数论);
代码:
#include<bits/stdc++.h>#define MAX 100010char n[MAX];int main(){ int num; while(~scanf("%s",n)) { int len=strlen(n); num=len>=2?10*(n[len-2]-'0')+(n[len-1]-'0'):n[0]-'0'; if(num%4==0) printf("4\n"); else printf("0\n"); memset(n,'\0',sizeof(n)); } return 0;}
0 0
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