浙江13省赛B 拼凑砝码

题目

Break Standard Weight

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

给你两个砝码的重量x和y，现在可以把其中一个拆成两部分，再求用这3个砝码可以量出的全部重量数。明显的3个物品装背包，看容量是否可以到达，可是当做背包做时间好多 囧

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define N 110
#define INF 1<<30
int cmp(const void*a,const void*b)
{
    return *(int*)a-*(int*)b;
}
int max;
int hash[N][N][N],f[4][N*2],s[4],n;
void dp()
{
    int i,j,k,l,w;
    memset(f,0,sizeof(f));
    f[0][0]=1;
    int ans=0;
    for(i=1;i<=3;i++)
    {
        w=s[i];
        memcpy(f[i],f[i-1],sizeof(f[0]));
        for(j=n;j>0;j--)
        {
            if(j>=w)//能补，能差j-w
            {
                if(f[i-1][j-w])//补成j
                    f[i][j]=1;
                if(f[i-1][j]) //差j-w;
                    f[i][j-w]=1;
            }
            else
            {
                if(f[i-1][j])//差w-j
                    f[i][w-j]=1;
            }
        }
    }
    for(i=1;i<=n;i++)
    {
        if(f[3][i])
            ans++;
    }
    if(ans>max)
        max=ans;
}
int main()
{
    int i,j,k,l;
    scanf("%d",&l);
    int a,b,c,x,y;
    while(l--)
    {
        memset(hash,0,sizeof(hash));
        max=-INF;
        scanf("%d %d",&x,&y);
        n=x+y;
        s[0]=-INF;
        for(i=1;i<=x/2;i++)
        {
            s[1]=i;
            s[2]=x-i;
            s[3]=y;
            qsort(s,4,sizeof(int),cmp);
            if(!hash[s[1]][s[2]][s[3]])
            {
                hash[s[1]][s[2]][s[3]]=1;
                dp();
            }
        }
        for(i=1;i<=y/2;i++)
        {
            s[1]=i;
            s[2]=y-i;
            s[3]=x;
            qsort(s,4,sizeof(int),cmp);
            if(!hash[s[1]][s[2]][s[3]])
            {
                hash[s[1]][s[2]][s[3]]=1;
                dp();
            }
        }
        printf("%d\n",max);
    }
    return 0;
}