Codeforces 260 B. Fedya and Maths
2014-08-10 21:47
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题目链接:http://codeforces.com/contest/456/problem/B
解题报告:输入一个n,让你判断(1n + 2n + 3n + 4n) mod 5的结果是多少?注意n的范围很大很大 n (0 ≤ n ≤ 10105).
只要判断是否能被整除4就可以了,如果n能被4整除,则结果是4,如果不能,则结果是0
但n很长,不能直接mod,但只要判断最低的两位能不能被4整除就可以了,如果n只有一位就判断最低的一位。
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解题报告:输入一个n,让你判断(1n + 2n + 3n + 4n) mod 5的结果是多少?注意n的范围很大很大 n (0 ≤ n ≤ 10105).
只要判断是否能被整除4就可以了,如果n能被4整除,则结果是4,如果不能,则结果是0
但n很长,不能直接mod,但只要判断最低的两位能不能被4整除就可以了,如果n只有一位就判断最低的一位。
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<cmath> #include<deque> #include<queue> #include<set> #include<map> using namespace std; #define maxn 100005 char str[maxn]; int main() { while(scanf("%s",str)!=EOF) { int temp,len = strlen(str); if(len <= 1) temp = str[0] - '0'; else temp = 10 * (str[len - 2] - '0') + (str[len - 1] - '0'); printf(temp % 4? "0\n":"4\n"); } return 0; }
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