Hdu 1709 The Balance

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3673    Accepted Submission(s): 1457

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

 

Recommend

lcy

题意：给你n个砝码，求出1-所有砝码总重量之间不能用砝码称出的重量，输出不能称出的砝码个数和从小到打输出不能称出的重量

解题思路：

      母函数。

原本用背包做了一次，WA，不过没找出来哪里错了，没想到这次用母函数做，居然AC了Orz......

一直以为用背包做这种问题会比较简单。。。。

#include<stdio.h>

int num[101];
int c1[10001],c2[10001];

int main(void)
{
    int n;
//    freopen("d:\\in.txt","r",stdin);
    while(scanf("%d",&n)==1)
    {
        int i,j,k;
        int max=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&num[i]);
            max+=num[i];
        }
        for(i=1;i<=max;i++)
        {
            c1[i]=0;
            c2[i]=0;
        }
        c1[0]=1;
        c1[num[1]]=1;
        for(i=2;i<=n;i++)
        {
            for(j=0;j<=max;j++)
            {
                if(c1[j])
                {
                    c2[j]=1;
                    if(j+num[i]<=max)
                        c2[j+num[i]]=1;
                }
                if(j-num[i]>=0 && c1[j])
                    c2[j-num[i]]=1;
                if(num[i]-j>=0 && c1[j])
                    c2[num[i]-j]=1;
            }
            for(j=0;j<=max;j++)
            {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        int count=0;
        for(i=1;i<=max;i++)
        {
            if(!c1[i])
                count++;
        }
        printf("%d\n",count);
        for(i=1;i<=max;i++)
        {
            if(!c1[i])
            {
                printf("%d",i);
                if(count>1)
                    printf(" ");
                else
                    printf("\n");
                count--;
            }
        }
    }
    return 0;
}