首页 > 代码库 > 平方数

平方数

2024-07-16 18:13:10   221人阅读

平方数
Time Limit : 3000/1000ms (Java/Other) Memory Limit : 65535/32768K (Java/Other)
Total Submission(s) : 3 Accepted Submission(s) : 3

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

Problem Description

Square number is very popular in ACM/ICPC. Now here is a problem about square number again.
Let‘s consider such a kind of number called K-Omitted-Square-Number(K-OSN). N is a K-OSN if:
(1) It is a square number.
(2) Its last digit is not 0.
(3) It‘s not less than 10^K.
(4) If its last K digits are omitted, it‘s still a square number.
Now, given an even number K, you have to find the largest K-OSN.

Input

There are multiple test cases in this problem.
The first line will contain a single positive integer T(T<=20) indicating the number of test cases. Each test case will be a single line containing an even number K(2<=K<=200).

Output

For each test case, print a single line containing the largest K-OSN. In case that it will be very large, you just need to print the number module 2009. If there are no K-OSN, or if the largest K-OSN doesn‘t exist, print "Oops!"(without quotes).

Sample Input

1
4

Sample Output

197

Author

HYNU
 
//数学题：  假定 m=a*10^k/2+b  (a,b未知）  N=m*m=a*a*10^k+b*b+2*a*b*10^k/2      
那么 b*b+2*a*b*10^k/2<10^k     因为要使m最大，那么b取最小，a取越大，所以b等于1  带入不等式然后对a取整就可以求得a，b。 a<(10^k-1)/(2*10^k/2)
但是我们可以发现，当k=2时，a=4  k=4  , a=49  k=6  a=499 ......所以我们可以构造出m，然后直接求出n.
#include<stdio.h>
const int MAXK=200;     //最大的K
const int MODNUM=2009;  //用于取模的数
int main()
{
    int tn,i;           //tn为数据组数
    int k,ans;          //题目输入的K
    int digit[MAXK+10],dn; //用于存放答案的数组和数组的长度
    scanf("%d",&tn);
    while(tn--)
    {
        scanf("%d",&k);
        dn=0; digit[++dn]=4;    //根据规律来构造答案
        for(i=1;i<k/2;i++) digit[++dn]=9;
        for(i=1;i<k/2;i++) digit[++dn]=0;
        digit[++dn]=1;
       // for(i=1;i<=dn;i++)
            //printf("%d ",digit[i]);
       // printf("\n");
        for(ans=0,i=1;i<=dn;i++) ans=(ans*10+digit[i])%MODNUM;
        //printf("%d\n",ans);
        ans=(ans*ans)%MODNUM;
        printf("%d\n",ans);
    }
    return 0;
}

 


联系
我们