CF456B Fedya and Maths 找规律

http://codeforces.com/contest/456/problem/B

CF#260 div2 B Fedya and Maths

Codeforces Round #260

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 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:

题解：

打表找规律，发现输入是4的倍数就出4，否则出0。

而4的倍数只用看最后两位，怕了。

 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 using namespace std;
 #define ll long long
 #define usint unsigned int
 #define mz(array) memset(array, 0, sizeof(array))
 #define minf(array) memset(array, 0x3f, sizeof(array))
 #define REP(i,n) for(i=0;i<(n);i++)
 #define FOR(i,x,n) for(i=(x);i<=(n);i++)
 #define RD(x) scanf("%d",&x)
 #define RD2(x,y) scanf("%d%d",&x,&y)
 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
 #define WN(x) printf("%d\n",x);
 #define RE  freopen("D.in","r",stdin)
 #define WE  freopen("1biao.out","w",stdout)

 int main() {
     //RE;
     ll n=;
     char c,prec,preprec;
     while(scanf("%c",&c)!=EOF && c>='' && c<='') {
         preprec=prec;
         prec=c;
         n++;
     }
     //printf("%c,%c,%c\n",c,prec,preprec);
     ll t;
     if(n>) t=(preprec-'')*+(prec-'');
     else t=prec-'';
     if(t%==)puts("");
     else puts("");
     return ;
 }