Pagodas 等差数列

标签：nal   http   integer   ase   element   inpu   tac   rebuild   from   

InputThe first line contains an integer t (1≤t≤500)t (1≤t≤500) which is the number of test cases. 
For each test case, the first line provides the positive integer n (2≤n≤20000)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\n", cas);
        else
            printf("Case #%d: Yuwgna\n", cas);
    }
}

Pagodas 等差数列

标签：nal   http   integer   ase   element   inpu   tac   rebuild   from   

原文地址：http://www.cnblogs.com/joeylee97/p/7395406.html