HDOJ 5610 Baby Ming and Weight lifting(枚举)
2016-01-23 22:12
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Total Submission(s): 68 Accepted Submission(s): 32
[align=left]Problem Description[/align]
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)
[align=left]Input[/align]
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
[align=left]Output[/align]
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)
[align=left]Sample Input[/align]
2
1 2 6
1 4 5
[align=left]Sample Output[/align]
2 2
Impossible
题意:有两个杠铃片,重量分别为a,b。要想使杠铃总重量为C,需要a,b的个数分别为多少?要求输出的num_a,num_b使得num_a+num_b最小。
题解:杠铃左右两边重量相等的,只考虑一边就可以了,枚举使得num_a*a+num_b*b=C/2的最小的num_a,num_b。最后输出记得乘以2就行了。
代码如下:
Baby Ming and Weight lifting
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 68 Accepted Submission(s): 32
[align=left]Problem Description[/align]
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)
[align=left]Input[/align]
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
[align=left]Output[/align]
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)
[align=left]Sample Input[/align]
2
1 2 6
1 4 5
[align=left]Sample Output[/align]
2 2
Impossible
题意:有两个杠铃片,重量分别为a,b。要想使杠铃总重量为C,需要a,b的个数分别为多少?要求输出的num_a,num_b使得num_a+num_b最小。
题解:杠铃左右两边重量相等的,只考虑一边就可以了,枚举使得num_a*a+num_b*b=C/2的最小的num_a,num_b。最后输出记得乘以2就行了。
代码如下:
#include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; #define INF 0x3f3f3f int main() { int t,a,b,C; int n,cnt_a,cnt_b,i,j; scanf("%d",&t); while(t--) { scanf("%d%d%d",&a,&b,&C); cnt_a=INF; cnt_b=INF; for(i=0;;++i) { for(j=0;;++j) { if((a*i+b*j)*2>C) break; if((a*i+b*j)*2==C) { if(i+j<cnt_a+cnt_b) { cnt_a=i; cnt_b=j; } } } if(a*i>C) break; } if(cnt_a==INF&&cnt_b==INF) printf("Impossible\n"); else printf("%d %d\n",cnt_a*2,cnt_b*2); } return 0; }
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