GDUT_寒假训练题解报告_数论专题_个人题解报告——题目:B - Fedya and Maths
题目:
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.
Examples
Input
4
Output
4
Input
124356983594583453458888889
Output
0
这个一看,是费马小的基本运用:
a^(P-1)==1(mod P)
这个东西就能消去所有的(1,2,3,4)^4;
然后就是读入数据,取最后两个,为什么是最后两个呢,因为100是4的倍数,所以大于100的都没了,取N为最后两位,比如123444558,N就是58,然后N%=5就可以了,注意特判处理一下输入的n为个位数的情况:
完整代码:
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <climits>
#include <queue>
#include <stack>
using namespace std;
#define INF 0x3f3f3f3f
#define ULL unsigned long long
char n[100010];
int main()
{
scanf("%s",n);
int length=strlen(n);
int N;
if(length>1)N=(n[length-2]-'0')*10+n[length-1]-'0';
else N=n[length-1]-'0';
N%=4;
int sum=0;
int k;
for(int time=1;time<5;time++)
{
k=1;
for(int time1=0;time1<N;time1++)
{
k*=time;
}
sum+=k;
}
printf("%d\n",sum%5);
return 0;
}
来源:CSDN
作者:DevourPower
链接:https://blog.csdn.net/DevourPower/article/details/103986916