The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3249    Accepted Submission(s): 1294


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
Recommend lcy 題目型別：普通型母函式 題目描述：給你n個砝碼，一個天平，砝碼總共重s.問在[1,s]的這個區間裡，哪個重量稱不出來。 題目分析：因為天平左右都能放砝碼，所以可以這樣。往左面放認為是正，往右面放認為是負。這樣每個砝碼的母函式相當於 x^-v[i] 1 x ^ v[i]. 程式碼如下：

#include <stdio.h>
#include <memory.h>
#define N 10001
#define M 101

int r[N];
int fr[N];
int t[N];
int ft[N];
int v[M];
int n,s;

int jd(int x){
    if(x < 0){
        x = -x;
    }
    return x;
}

void init(){
    int i,j,k;
    memset(r,0,sizeof(r));
    memset(fr,0,sizeof(fr));
    memset(t,0,sizeof(t));
    memset(ft,0,sizeof(ft));
    r[0] = 1,r[v[0]] = 1,fr[v[0]] = 1;

    for( k = 1; k < n; k++){
        for( i = 0; i <= s; i++){
            for( j = -v[k]; j <= v[k]; j += v[k]) {
                if( r[i] != 0) {
                    if(i+j < 0){
                        ft[jd(i+j)] = 1;
                    } else {
                        t[i+j] = 1;
                    }
                }
            }
        }
        for( i = 1; i <= s; i++){
            for( j = -v[k]; j <= v[k]; j += v[k]) {
                if( fr[i] != 0){
                    if( -i + j < 0){
                        ft[jd(-i+j)] = 1;
                    } else {
                        t[-i+j] = 1;
                    }
                }
            }
        }

        for( i = 0; i <= s ;i++){
            r[i] = t[i];
            t[i] = 0;
            fr[i] = ft[i];
            ft[i] = 0;
        }
    }
}
void result(){
    int i,sum = 0;
    for( i = 0; i <= s; i++){
        if(r[i] == 0){
            sum++;
        }
    }
    printf("%d\n",sum);
    if( sum != 0) {
        int flag = 0;
        for( i = 0; i <= s; i++){
            if(r[i] == 0) {
                if( flag == 0){
                    flag = 1;
                } else {
                    printf(" ");
                }
                printf("%d",i);
            }
        }
        printf("\n");
    }
}


int main()
{
    while(scanf("%d",&n) != EOF){
        s = 0;
        int i;
        for( i = 0; i < n; i++){
            scanf("%d",v+i);
            s += v[i];
        }
        init();
        result();
    }

    return 0;
}