hdu 1709(普通型母函数)

最新推荐文章于 2019-10-24 19:27:34 发布

jiakai0419

最新推荐文章于 2019-10-24 19:27:34 发布

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分类专栏： ACM/ICPC 文章标签： input each integer output

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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

Sample Output

Source

HDU 2007-Spring Programming Contest

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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;
}