BNU29378-水博弈

来源：互联网发布：如何查找快递行业数据编辑：程序博客网时间：2024/05/22 09:06

题目：题目链接

题意：如题，就是判断长和宽的可以变成的偶数的个数

代码：

#include <iostream>#include <cstdio>#include <string>#include <string.h>#include <map>#include <vector>#include <cstdlib>#include <algorithm>#include <cmath>#include <queue>#include <set>#include <stack>#include <functional>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cassert>#include <bitset>#include <stack>#include <ctime>#include <list>#define INF 0x7fffffff#define max3(a,b,c) (max(a,b)>c?max(a,b):c)#define min3(a,b,c) (min(a,b)<c?min(a,b):c)#define mem(a,b) memset(a,b,sizeof(a))using namespace std;int QuickMod(int  a,int b,int n){    int r = 1;    while(b)    {        if(b&1)            r = (r*a)%n;        a = (a*a)%n;        b >>= 1;    }    return r;}int main(){    int t;    scanf("%d", &t);    int c, l;    while(t--)    {        scanf("%d%d", &l, &c);        int f1 = 0, f2 = 0;        while(l%2==0)        {            f1++;            l /= 2;        }        while(c%2==0)        {            f2++;            c /= 2;        }        if((f1+f2)%2 == 0)printf("Adivon prevails\n");        else printf("Adidas loses\n");    }    return 0;}