HDOJ 5610-Baby Ming and Weight lifting

Myth丶恋晨 2022-08-19 15:15 14阅读 0赞

#  #

# Baby Ming and Weight lifting #

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 222 Accepted Submission(s): 91

Problem Description

Baby Ming is fond of weight lifting. He has a barbell pole(the weight of which can be ignored) and two different kinds of barbell disks(the weight of which are respectively a![Image 1][] and b![Image 1][]), the amount of each one being infinite.  
  
Baby Ming prepare to use this two kinds of barbell disks to make up a new one weighted C![Image 1][](the barbell must be balanced), he want to know how to do it.

![Center][]

Input

In the first line contains a single positive integer T![Image 1][], indicating number of test case.  
  
For each test case:  
  
There are three positive integer a,b![Image 1][], and C![Image 1][].  
  
1≤T≤1000,0<a,b,C≤1000,a≠b![Image 1][]

Output

For each test case, if the barbell weighted C![Image 1][] can’t be made up, print Impossible.  
  
Otherwise, print two numbers to indicating the numbers of a![Image 1][] and b![Image 1][] barbell disks are needed. (If there are more than one answer, print the answer with minimum a\+b![Image 1][])

Sample Input

2
    1 2 6
    1 4 5

Sample Output

2 2
    Impossible

Source

[BestCoder Round \#69 (div.2) ][BestCoder Round _69 _div.2_]

题目大意：

# 就是给出a，b这两个杠铃盘的重量，给出C这个他想要举起这个杠铃的重量，在保证左右两边平衡的状态下，用最少的盘。 #

解题思路：

暴力即可，注意一下输出为 **Impossible的情况。**

#include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #define INF 0x3f3f3f3f 
    using namespace std;
    int main()
    {
        int T;
        scanf("%d",&T);
        while(T--)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            if(c%2!=0)
            {
                printf("Impossible\n");
            } 
            else
            {
                int ni=INF,wo1=INF;
                int ans1,ans2;
                int da,xiao;
                da=max(a,b);
                xiao=min(a,b);
                int uu=c/min(a,b);
                for(int i=0;i<=uu;i+=2)
                {
                    if(((c-(xiao*i))/da)%2==0)
                    {
                        ans1=i;
                        ans2=(c-(xiao*i))/da;
                        if(ans1*xiao+ans2*da==c)
                        {
                            ni=ans1;
                            wo1=ans2;
                            break;    
                        }
                    }
                }
                if(ni==INF||wo1==INF)
                {
                    printf("Impossible\n");    
                }
                else
                {
                    if(a<b)
                    printf("%d %d\n",ans1,ans2);
                    else
                    printf("%d %d\n",ans2,ans1);
                }
            }
        }
        return 0;
    }

[Image 1]: 
[Center]: /images/20220731/ec6487c1ea704039890f1e4f1c85701f.png
[BestCoder Round _69 _div.2_]: http://acm.hdu.edu.cn/search.php?field=problem&key=BestCoder+Round+%2369+%28div.2%29&source=1&searchmode=source