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  • Pagodas 等差数列

    nn pagodas were standing erect in Hong Jue Si between the Niushou Mountain and the Yuntai Mountain, labelled from 11 to nn. However, only two of them (labelled aa and bb, where 1abn1≤a≠b≤n) withstood the test of time. 

    Two monks, Yuwgna and Iaka, decide to make glories great again. They take turns to build pagodas and Yuwgna takes first. For each turn, one can rebuild a new pagodas labelled i (i{a,b} and 1in)i (i∉{a,b} and 1≤i≤n) if there exist two pagodas standing erect, labelled jj and kkrespectively, such that i=j+ki=j+k or i=jki=j−k. Each pagoda can not be rebuilt twice. 

    This is a game for them. The monk who can not rebuild a new pagoda will lose the game.

    InputThe first line contains an integer t (1t500)t (1≤t≤500) which is the number of test cases. 
    For each test case, the first line provides the positive integer n (2n20000)n (2≤n≤20000) and two different integers aa and bb.
    OutputFor each test case, output the winner (``Yuwgna" or ``Iaka"). Both of them will make the best possible decision each time.Sample Input

    16
    2 1 2
    3 1 3
    67 1 2
    100 1 2
    8 6 8
    9 6 8
    10 6 8
    11 6 8
    12 6 8
    13 6 8
    14 6 8
    15 6 8
    16 6 8
    1314 6 8
    1994 1 13
    1994 7 12

    Sample Output

    Case #1: Iaka
    Case #2: Yuwgna
    Case #3: Yuwgna
    Case #4: Iaka
    Case #5: Iaka
    Case #6: Iaka
    Case #7: Yuwgna
    Case #8: Yuwgna
    Case #9: Iaka
    Case #10: Iaka
    Case #11: Yuwgna
    Case #12: Yuwgna
    Case #13: Iaka
    Case #14: Yuwgna
    Case #15: Iaka
    Case #16: Iaka

    n个数字相加或者相减得出的数列是gcd(a,b)的等差数列,求出有几项就OK了
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<sstream>
    #include<algorithm>
    #include<queue>
    #include<vector>
    #include<cmath>
    #include<map>
    #include<stack>
    #include<fstream>
    #include<set>
    #include<memory>
    #include<bitset>
    #include<string>
    #include<functional>
    using namespace std;
    typedef long long LL;
    
    int T, a, b, n;
    int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        else
            return gcd(b, a%b);
    }
    int main()
    {
        scanf("%d", &T);
        for(int cas = 1;cas<=T;cas++)
        {
            scanf("%d%d%d",  &n,&a ,&b);
            if (!((n / gcd(a, b)) & 1))
                printf("Case #%d: Iaka
    ", cas);
            else
                printf("Case #%d: Yuwgna
    ", cas);
        }
    }
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  • 原文地址:https://www.cnblogs.com/joeylee97/p/7395406.html
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