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  • ZOJ Problem Set

    http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3706
    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 4 and 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
    给出两个数,把其中一个数分割,分为a,b,c
    这三个数相加减,看看能得到多少不同的数
    0不算,
    暴力

    #include<stdio.h>
    #include<math.h>
    #include<iostream>
    #include<algorithm>
    #include<set>
    using namespace std;
    int main()
    {
        int t;
        scanf("%d",&t);
        while(t--)
        {
            int x,y;
            int sum=0;
            scanf("%d%d",&x,&y);
            for(int i=0; i<=x/2; i++)
            {
                int a=i;
                int b=x-i;
                int c=y;
                set<int>s;
                s.insert(a);
                s.insert(b);
                s.insert(c);
                s.insert(a+b);
                s.insert(b+c);
                s.insert(a+c);
                s.insert(abs(a-b));
                s.insert(abs(a-c));
                s.insert(abs(b-c));
                s.insert(a+b+c);
                s.insert(abs(a+b-c));
                s.insert(abs(a-b+c));
                s.insert(abs(a-b-c));
                s.insert(abs(-a+b+c));
                s.insert(abs(-a+b-c));
                s.insert(abs(-a-b+c));
                s.insert(abs(-a-b-c));
                set<int>::iterator it=s.begin();
                int len=s.size();
                if(*it==0)//0不算
                    len--;
                if(len>sum)
                    sum=len;
            }
            for(int i=0; i<=y/2; i++)
            {
                int a=i;
                int b=y-i;
                int c=x;
                set<int>s;
                s.insert(a);
                s.insert(b);
                s.insert(c);
                s.insert(a+b);
                s.insert(b+c);
                s.insert(a+c);
                s.insert(abs(a-b));
                s.insert(abs(a-c));
                s.insert(abs(b-c));
                s.insert(a+b+c);
                s.insert(abs(a+b-c));
                s.insert(abs(a-b+c));
                s.insert(abs(a-b-c));
                s.insert(abs(-a+b+c));
                s.insert(abs(-a+b-c));
                s.insert(abs(-a-b+c));
                s.insert(abs(-a-b-c));
                set<int>::iterator it=s.begin();
                int len=s.size();
                if(*it==0)
                    len--;
                if(len>sum)
                    sum=len;
            }
            printf("%d
    ",sum);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zxy160/p/7215128.html
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