CSU 规律题

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Description

Input

Output

Sample Input

32 2 22 1 310 1000 24

Sample Output

HINT

Source

根据题意,看一个数可以分多少次才可以到达 a和b的值

#include <iostream>#include <algorithm>#include <cstdio>#include<cmath>#include <cstring>#include <cstdlib>#include <map>#define inf 0x3f3f3f3f#define Max 20010using namespace std;int main(){   int T;   long long a,b,n,sum=0;   while(~scanf("%d",&T))   {       while(T--)       {           scanf("%lld%lld%lld",&n,&a,&b);           if(a>b)            swap(a,b);           sum=pow(2,n);           long long num=0;           while(1)           {               if(a%sum==0)               {                   break;               }               else               {                   num++;                   sum/=2;               }           }           printf("%lld\n",num);       }   }}

1535: Pizza voting

Time Limit: 5 Sec Memory Limit: 128 MB
Submit: 522 Solved: 192
[Submit][Status][Web Board]

Description

Input

Output

Sample Input

5 2500 Margherita600 Salami700 Hawai800 Speciale900 Doener

Sample Output

YES

HINT

Source

//将每种情况都手写一遍,查找规律题意是A喜欢能量最少的,B喜欢能量最大的.所以每次A先选能量最大的排除掉,然后B选能量最小的除掉,然后该我聪明的选一个排除,多次循环后,使你选的号码保留到最后.#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <cstdlib>using namespace std;int main(){    int n,m;    char s[1000];    int a;    while(~scanf("%d%d",&n,&m))    {        for(int i=0;i<n;i++)        {            scanf("%d%s",&a,s);        }        if(n/3<m&&m<=(n/3+(n-1)/3)+1)            printf("YES\n");        else            printf("NO\n");    }    return 0;}

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