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

1 2 4

9 2 1

Sample Output

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

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                   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|1typedef long long LL;typedef pair
                   
                    pii;#define X first#define Y secondconst 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();}

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转载于:https://www.cnblogs.com/hlmark/p/4204620.html

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