zoj 3706 Break Standard Weight(数学题)
2014-03-22 09:41
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题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5009
Break Standard Weight
Time Limit: 2 Seconds Memory Limit: 65536 KB
The balance was the first mass measuring instrument invented. In its traditional form, it consists of a pivoted horizontal lever of equal length arms, called the beam,
with a weighing pan, also called scale, suspended from each arm (which is the origin of the originally plural term "scales" for a weighing instrument). The unknown mass is placed in one pan, and standard masses are added to this or the other pan until the
beam is as close to equilibrium as possible. The standard weights used with balances are usually labeled in mass units, which are positive integers.
With some standard weights, we can measure several special masses object exactly, whose weight are also positive integers in mass units. For example, with two standard weights 1 and 5,
we can measure the object with mass 1, 4, 5 or 6 exactly.
In the beginning of this problem, there are 2 standard weights, which masses are x and y. You have to choose a standard weight to break it into 2 parts,
whose weights are also positive integers in mass units. We assume that there is no mass lost. For example, the origin standard weights are 4 and 9, if you break the second one into 4and 5,
you could measure 7 special masses, which are 1, 3, 4, 5, 8, 9, 13. While if you break the first one into 1 and 3, you could measure 13 special masses, which are 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13! Your task is to find out the maximum number of possible special masses.
2 ≤ x, y ≤ 100
Author: YU, Zhi
Contest: The 10th Zhejiang Provincial Collegiate Programming Contest
题意很简单: 就是给你两个数X,Y! 让你把其中一个数分为两部分!a+b=X 或 a+b = Y;!然后 找出怎样拆分才能让这三个数通过加减运算有最后的不同答案!并输出个数!
PS:
总共也就三个数,三个数的加减运算的答案不会太多!我们可以一一列举出来!
代码如下:
题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5009
Break Standard Weight
Time Limit: 2 Seconds Memory Limit: 65536 KB
The balance was the first mass measuring instrument invented. In its traditional form, it consists of a pivoted horizontal lever of equal length arms, called the beam,
with a weighing pan, also called scale, suspended from each arm (which is the origin of the originally plural term "scales" for a weighing instrument). The unknown mass is placed in one pan, and standard masses are added to this or the other pan until the
beam is as close to equilibrium as possible. The standard weights used with balances are usually labeled in mass units, which are positive integers.
With some standard weights, we can measure several special masses object exactly, whose weight are also positive integers in mass units. For example, with two standard weights 1 and 5,
we can measure the object with mass 1, 4, 5 or 6 exactly.
In the beginning of this problem, there are 2 standard weights, which masses are x and y. You have to choose a standard weight to break it into 2 parts,
whose weights are also positive integers in mass units. We assume that there is no mass lost. For example, the origin standard weights are 4 and 9, if you break the second one into 4and 5,
you could measure 7 special masses, which are 1, 3, 4, 5, 8, 9, 13. While if you break the first one into 1 and 3, you could measure 13 special masses, which are 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13! Your task is to find out the maximum number of possible special masses.
Input
There are multiple test cases. The first line of input is an integer T < 500 indicating the number of test cases. Each test case contains 2 integers x and y.2 ≤ x, y ≤ 100
Output
For each test case, output the maximum number of possible special masses.Sample Input
2 4 9 10 10
Sample Output
13 9
Author: YU, Zhi
Contest: The 10th Zhejiang Provincial Collegiate Programming Contest
题意很简单: 就是给你两个数X,Y! 让你把其中一个数分为两部分!a+b=X 或 a+b = Y;!然后 找出怎样拆分才能让这三个数通过加减运算有最后的不同答案!并输出个数!
PS:
总共也就三个数,三个数的加减运算的答案不会太多!我们可以一一列举出来!
代码如下:
#include<cstdio> #include<iostream> using namespace std; int Find( int x , int y, int z ) { int count = 0; int s[20]; int i, j; s[0] = x ; s[1] = y ; s[2] = z ; s[3] = x + y ; s[4] = x + z ; s[5] = y + z ; s[6] = x - y ; s[7] = y - z ; s[8] = x - z ; s[9] = y - x ; s[10] = z - x ; s[11] = z - y ; s[12] = x - y - z ; s[13] = x - y + z ; s[14] = y - x - z ; s[15] = y - x + z ; s[16] = y - z + x ; s[17] = z - x + y ; s[18] = z - x - y ; s[19] = x + y + z ; for( i = 0 ; i < 20 ; i++) { for( j = i + 1 ; j < 20 ; j++ ) { if(s[i] > 0 && s[j] > 0 && s[i] == s[j]) { s[j] = -1 ; } } } for( i = 0 ; i < 20 ; i++ ) { if( s[i] > 0 ) count++; } return count ; } int main() { int T, a, b; int i, j, max, tmp; while( cin >> T) { while(T--) { max = 0; cin >> a >> b; for( i = 1 ; i < a/2+1 ; i++ ) { j = a - i ; tmp = Find(i , j , b); if( max < tmp ) max = tmp; } for(i = 1 ; i< b/2+1 ; i++ ) { j = b - i ; tmp = Find(i , j , a); if( max < tmp) max = tmp; } cout<< max <<endl; } } return 0 ; }
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