SDUST 2844-Mineral Water(数学)-白红宇

SDUST 2844-Mineral Water(数学)

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发布时间：2019-06-24

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Mineral Water

Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^

题目描写叙述

Laoshan mineral water is one of famous well-known mineral water, Tyrant(means”Tu Hao”) Chierush liked to drink this mineral water very much. He said that it is such a natural taste, but this is not the focus. One day Chierush went to the campus supermarket to buy n bottles of Laoshan mineral water, the boss see that he is an acquaintance, so he gave him a copy of preferential policies - he can change every m empty bottles into one bottle mineral water. Your task is to help Chierush calculate how many bottles of mineral water he can drink?

输入

The input consists of multiple test cases. The first line contains an integer T means the number of test cases. Each test case consists of one line with two numbers represent n and m. (T<=100, 2<=n,m<=2*10^9)

输出

For each test case, output one line, including one number that represents the answer.

演示样例输入

22 25 4

演示样例输出

貌似山东某届的省赛题？也是够水的了

题意 ：n瓶水，m个空瓶能够换一瓶水。问最多能够喝多少瓶水。

注意用LL 然后没什么了。先喝掉m的整数倍。然后剩下的n%m瓶加上能够换到水的瓶数，这时候把这个数看成最開始那个n就能够了。循环搞定。

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              
               #include 
               
                #include 
                #include 
                 
                  #define ll long longusing namespace std;const int INF=0x3f3f3f3f;int main(){ ll n,m;int t; scanf("%d",&t); while(t--) { scanf("%lld%lld",&n,&m); ll ans=0; while(n>=m) { ans+=(n-n%m); ll tem=n/m; n%=m; n+=tem; } ans+=n; printf("%lld\n",ans); } return 0;}

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