hdu 1709(普通型母函数)

本文介绍了如何通过给定数量和质量的砝码,利用天平来测量特定范围内的任意质量。详细阐述了算法设计、实现步骤及样例输入输出,包括初始化、计算不可测量的质量范围并输出结果。

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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
 

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题目类型:普通型母函数

题目描述:给你n个砝码,一个天平,砝码总共重s.问在[1,s]的这个区间里,哪个重量称不出来。

题目分析:因为天平左右都能放砝码,所以可以这样。往左面放认为是正,往右面放认为是负。这样每个砝码的母函数相当于 x^-v[i] 1 x ^ v[i].


代码如下:


#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;
}


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