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  • LightOJ

    Time Limit: 2000MS   Memory Limit: 32768KB   64bit IO Format: %lld & %llu

    Status

    Description

    Ekka and his friend Dokka decided to buy a cake. They both love cakes and that's why they want to share the cake after buying it. As the name suggested that Ekka is very fond of odd numbers and Dokka is very fond of even numbers, they want to divide the cake such that Ekka gets a share of N square centimeters and Dokka gets a share of M square centimeters where N is odd and M is even. Both N and M are positive integers.

    They want to divide the cake such that N * M = W, where W is the dashing factor set by them. Now you know their dashing factor, you have to find whether they can buy the desired cake or not.

    Input

    Input starts with an integer T (≤ 10000), denoting the number of test cases.

    Each case contains an integer W (2 ≤ W < 263). And W will not be a power of 2.

    Output

    For each case, print the case number first. After that print "Impossible" if they can't buy their desired cake. If they can buy such a cake, you have to print N and M. If there are multiple solutions, then print the result where M is as small as possible.

    Sample Input

    3

    10

    5

    12

    Sample Output

    Case 1: 5 2

    Case 2: Impossible

    Case 3: 3 4

    Source

    Problem Setter: Muhammad Rifayat Samee
    Special Thanks: Jane Alam Jan

    Status

    奇数不会拆分成 奇 * 偶 形式;

    #include <cstdio>
    int main()
    {
        int t;
        scanf("%d", &t);
        int Q=1;
        while(t--)
        {
            long long w;
            scanf("%lld", &w);
            if(w&1)
                {
                    printf("Case %d: Impossible
    ", Q++);
                }
            else
            {
                long long i;
                for(i=2; i<= w; i+=2)
                {
                    long long extra=w/i;   //注意是取整,  不一定可以整除 ; 
                    if(extra*i==w && extra&1) 
                    {
                        break;
                    }
                }
                printf("Case %d: %lld %lld
    ", Q++, w/i, i);
            }
        }
        return 0;    
    }

     

     

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  • 原文地址:https://www.cnblogs.com/soTired/p/5330821.html
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