HDU 1709 The Balance( DP )

The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5956    Accepted Submission(s): 2427

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

 

 

一条挺有意思的递推。

就是问给出n个重量为wi的砝码。

问你不能称出的重量有哪些。

bool  dp[i][j] 表示前i个砝码， 能否称出重量 j 。

 

初始化dp[0][0] = true ;

对于一个 dp[i-1][j] = true , 必然有 dp[i][j+a[i]] = true , dp[i][abs(j-a[i])] = true 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <algorithm>
using namespace std;
#define root 1,n,1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define lr rt<<1
#define rr rt<<1|1
typedef long long LL;
typedef pair<int,int>pii;
#define X first
#define Y second
const int oo = 1e9+7;
const double PI = acos(-1.0);
const double eps = 1e-6 ;
const int N = 10500;
const int mod = 1e9+7;
bool dp[105][N];
int n , m , a[N] , b[N] , tot ;
void init() {
    memset( dp , false ,sizeof dp );
    dp[0][0] = true ;
    for( int i = 1 ; i <= n ; ++i ){
        for( int j = 0 ; j <= tot ; ++j ){
            if( dp[i-1][j] ) dp[i][j] = dp[i-1][j];
            if( dp[i-1][j] ){
                dp[i][j+a[i]] = true ; dp[i][abs(j-a[i])] = true ;
            }
        }
    //    for( int j = 0 ; j <= tot ; ++j ) cout << dp[i][j] << ' '; cout << endl ;
    }
}
void Run() {
    tot = 0 ;
    for( int i = 1 ; i <= n ; ++i ) 
        cin >> a[i] , tot += a[i];
    init(); int cnt = 0 ;    
    for( int i = 1 ; i <= tot ; ++i ) if( !dp[n][i] ){
        b[cnt++] = i ;
    }
    cout << cnt << endl ;
    for( int i = 0 ; i < cnt ; ++i ) 
        cout << b[i] << (i+1==cnt?"\n":" ");

}
int main()
{
//    freopen("in.txt","r",stdin);
    ios::sync_with_stdio(false);
    while( cin >> n ) Run();
}