URAL 1295. Crazy Notions(數學啊 & 找規律)
來源：程序員人生 發布時間：2015-04-08 08:35:27 閱讀次數：3506次
題目鏈接：http://acm.timus.ru/problem.aspx?space=1&num=1295
1295. Crazy Notions
Time limit: 0.5 second
Memory limit: 64 MB
For five days robot-loader JK546L54p has been buried under the thick layer of the Sibelian plutonium slag. The terrible strike of the atmospheric electricity has led to the depressurization of the robot’s
fuel elements. Who will examine this heap of fused, broken metal here, where there is no any robot technician even at distance of a hundred parsecs? Robot-commissar even did not try to investigate what happened with JK546L54p. He ordered to throw him out into
dumps and that is all. Nobody noticed that positron brains of JK546L54p were still working. If only the robopsychologist was here with JK546L54p! Of course, he would be killed with the hard gamma radiation in a moment, but… If he attached the visualizer of
thoughts to the fused connectors of JK546L54p! He would see the strange performance. Robot was creating! No, I am not joking. He was investigating. Semi casual objects arose in his mind, and he examined them. Crazy properties, crazy theorems.
Besides, here is an example. Let’s take an expression 1n+2n+3n+4n.
How much zeros does its decimal notation end with? JK546L54p solved this problem, and you, student, could you?
Input
The only line contains an integer n (1 ≤ n ≤ 300000).
Output
Output the number of zeroes the decimal notation of 1n+2n+3n+4n ends
with.
Samples
代碼以下（找規律）：
#include <cstdio>
#define LL __int64
//int main()
//{
// int n;
// for(int i = 1; i <= 32; i++)
// {
// LL ans1 = 1;
// LL ans2 = 1;
// LL ans3 = 1;
// LL ans4 = 1;
// for(int j = 1; j <= i; j++)
// {
// ans2*=2;
// ans3*=3;
// ans4*=4;
// }
// int cnt = 0;
// LL ans = ans1+ans2+ans3+ans4;
// while(ans)
// {
// if(ans%10 == 0)
// {
// cnt++;
// ans/=10;
// }
// else
// {
// break;
// }
// }
// printf("%d:: ans:%d
",i,cnt);
// }
// return 0;
//}
int main()
{
int n;
int a[30]= {1,1,2,0,2,1,2,0,1,1,2,0,1,1,2,0,1,1,2,0};
while(~scanf("%d",&n))
{
printf("%d
",a[(n⑴)%20]);
}
return 0;
// while(~scanf("%d",&n))
// {
// if(n%4 == 0)
// {
// printf("0
");
// }
// else if((n%4==1 &&n%5==0) || n%4==3)
// {
// printf("2
");
// }
// else
// {
// printf("1
");
// }
// }
return 0;
}
貼1發他人的：
#include <stdio.h>
#include <iostream>
#include <algorithm>
using namespace std;
long long mod=100000;
long long quickmulti(long long m,long long n)//2分快速冪
{
long long ans=1;
long long i;
while(n)
{
if(n&1)
ans=(m*ans)%mod;
m=(m*m)%mod;
n>>=1;
}
return ans;
}
int main()
{
long long n;
while(scanf("%lld",&n)!=EOF)
{
long long ans=1;
ans+=quickmulti(2,n);
ans%=mod;
ans+=quickmulti(3,n);
ans%=mod;
ans+=quickmulti(4,n);
ans%=mod;
long long tem=ans;
long long tt=0;
while(tem%10==0)
{
tem/=10;
tt++;
}
printf("%lld
",tt);
}
return 0;
}
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