HDU1709母函数

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7274    Accepted Submission(s): 3006

Problem Description

Now you are asked to measure a dose of medicine with a balance and a number of weights. Certainly it is not always achievable. So you should find out the qualities which cannot be measured from the range [1,S]. S is the total quality of all the weights.

 

Input

The input consists of multiple test cases, and each case begins with a single positive integer N (1<=N<=100) on a line by itself indicating the number of weights you have. Followed by N integers Ai (1<=i<=N), indicating the quality of each weight where 1<=Ai<=100.

 

Output

For each input set, you should first print a line specifying the number of qualities which cannot be measured. Then print another line which consists all the irrealizable qualities if the number is not zero.

 

Sample Input

   
   
   
   

    
    
    
    3
1 2 4
3
9 2 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0
2
4 5
   
   
   
   

 

Source

HDU 2007-Spring Programming Contest

/*
题意：找出用n个砝码称重不能表示的重量有哪些，从小到大输出。
注意：每种砝码既可以放左盘又可以放右盘，按照左物右码的规则，放在左盘的时候就取减号，放在右盘的时候就取加号。
*/
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>

using namespace std;

int c1[10010],c2[10010];
int ans[10010];
int a[111];
int sum;
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        sum = 0;
        for(int i = 1; i <= n; i ++)
        {
            scanf("%d",&a[i]);
            sum += a[i];
        }
        memset(c1,0,sizeof(c1));
        memset(c2,0,sizeof(c2));
        c1[0] = 1;///一定要注意这个初始化，非常有用，而且不会对最终正确的结果产生影响，但是没有它的话会错
        for(int i = 1; i <= n; i ++)
        {
            for(int j = 0; j <= sum; j ++)///j对应最大的重量
                ///(j+k<=sum)表明最大可取的情况不要大于sum,该括号中最大只能表示a[i],不能大于
                for(int k = 0; j + k <= sum && k <= a[i]; k += a[i])
                {
                    ///两括号中的砝码可能都是放在右边，或者放在两个盘子
                    ///注意，都放在左盘子的那就不是称重了
                    ///还有就是要对每种情况取绝对值
                    if(k >= j)
                        c2[k - j] += c1[j];
                    else
                        c2[j - k] += c1[j];
                    c2[k+j] += c1[j];
                }
            for(int j = 0; j <= sum; j ++)
            {
                c1[j] = c2[j];
                c2[j] = 0;
            }
        }
        memset(ans,0,sizeof(ans));
        int s = 0;///统计数量
        for(int i = 0; i <= sum; i ++)
            if(c1[i] == 0)
            {
                ans[s] = i;
                s ++;
            }
        printf("%d\n",s);
        if(s)
        {
            printf("%d",ans[0]);
            for(int i = 1; i < s; i ++)
                printf(" %d",ans[i]);
            printf("\n");
        }
    }
    return 0;
}