BestCoder Round #69 (div.2) Baby Ming and Weight lifting(hdu 5610)
Baby Ming and Weight lifting
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 681 Accepted Submission(s): 280
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 and b), 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(the barbell must be balanced), he want to know how to do it.
![](http://bestcoder.hdu.edu.cn/data/images/C664-1001-1.jpg)
Baby Ming prepare to use this two kinds of barbell disks to make up a new one weighted C(the barbell must be balanced), he want to know how to do it.
![](http://bestcoder.hdu.edu.cn/data/images/C664-1001-1.jpg)
Input
In the first line contains a single positive integer T, indicating number of test case.
For each test case:
There are three positive integer a,b, and C.
1≤T≤1000,0<a,b,C≤1000,a≠b
For each test case:
There are three positive integer a,b, and C.
1≤T≤1000,0<a,b,C≤1000,a≠b
Output
For each test case, if the barbell weighted C can’t be made up, print Impossible.
Otherwise, print two numbers to indicating the numbers of a and b barbell disks are needed. (If there are more than one answer, print the answer with minimum a+b)
Otherwise, print two numbers to indicating the numbers of a and b barbell disks are needed. (If there are more than one answer, print the answer with minimum a+b)
Sample Input
2
1 2 6
1 4 5
Sample Output
2 2
Impossible
Source
题意:有A B两种杠铃盘,杠铃杆的重量忽略,这两种盘都有无限个,现在让你用这两种盘来制作新的杠铃C问需要A B多少个(n,m) 如果有多种方案,取n+m最少的
注意:杠铃的两端必须重量相等,
分析1、c的值必须为偶数2、制作c必须使A的个数和B的个数都为偶数(可以为0)
题解:因为题目数据不大 所以直接枚举所有情况
#include<stdio.h> #include<string.h> #include<string> #include<math.h> #include<algorithm> #define LL long long #define PI atan(1.0)*4 #define DD doublea #define MAX 10100 #define mod 10007 using namespace std; int ans[MAX]; int main() { int n,m,j,i,t,k; int a,b,c,Min1,x,y; scanf("%d",&t); while(t--) { scanf("%d%d%d",&a,&b,&c); n=m=0; if(c&1) { printf("Impossible\n"); continue; } Min1=1000000;x=y=0; for(i=0;i<=1000;i+=2) { for(j=0;j<=1000;j+=2) { if(c==i*a+j*b) { if(i+j<Min1) { x=i;y=j; Min1=i+j; } } } } if(Min1>10000) printf("Impossible\n"); else printf("%d %d\n",x,y); } return 0; }