The Balance

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5463    Accepted Submission(s): 2214

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

题意：给定n个砝码的重量，总质量为sum，问在1~sum中有多少个重量不能被称出来。

题解：母函数的应用，需要注意sum的值。

版本一：171ms

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define maxn 10002

int arr[102], store[maxn];
bool c1[maxn], c2[maxn];

int main()
{
    int n, i, j, k, sum, count;
    while(scanf("%d", &n) != EOF){
        memset(c1, 0, sizeof(c1));
        memset(c2, 0, sizeof(c2));
        for(i = sum = 0; i < n; ++i){
            scanf("%d", arr + i);
            sum += arr[i];
        }
        c1[0] = c1[arr[0]] = 1;
        for(i = 1; i < n; ++i){
            for(j = 0; j <= sum; ++j)
                for(k = 0; k <= arr[i] && j + k <= sum; k += arr[i]){
                    c2[k + j] += c1[j];
                    c2[abs(k - j)] += c1[j];
                }
            for(k = 0; k <= sum; ++k){
                c1[k] = c2[k]; c2[k] = 0;
            }
        }
        for(count = 0, i = 1; i <= sum; ++i){
            if(!c1[i]) store[count++] = i;
        }
        printf("%d\n", count);
        if(count)
            for(i = 0; i < count; ++i)
                if(i != count - 1) printf("%d ", store[i]);
                else printf("%d\n", store[i]);
    }
    return 0;
}

版本二：78ms

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define maxn 10002

int arr[102], store[maxn];
bool c1[maxn], c2[maxn];

int main()
{
    int n, i, j, k, sum, count;
    while(scanf("%d", &n) != EOF){
        memset(c1, 0, sizeof(c1));
        memset(c2, 0, sizeof(c2));
        for(i = sum = 0; i < n; ++i)
            scanf("%d", arr + i);
        c1[0] = c1[sum = arr[0]] = 1;
        for(i = 1; i < n; ++i){
            for(j = 0; j <= sum; ++j)
                for(k = 0; k <= arr[i]; k += arr[i]){
                    c2[k + j] += c1[j];
                    c2[abs(k - j)] += c1[j];
                }
            sum += arr[i];
            for(k = 0; k <= sum; ++k){
                c1[k] = c2[k]; c2[k] = 0;
            }
        }
        for(count = 0, i = 1; i <= sum; ++i){
            if(!c1[i]) store[count++] = i;
        }
        printf("%d\n", count);
        if(count)
            for(i = 0; i < count; ++i)
                if(i != count - 1) printf("%d ", store[i]);
                else printf("%d\n", store[i]);
    }
    return 0;
}

HDU1709 The Balance 【母函数】,布布扣,bubuko.com