hdu 1709 母函数变形

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

#define MAX 10010

int c1[MAX], c2[MAX], num[110];

/* 题目大意： 给你n种砝码，从1到n中砝码的重量和其中不能称出多少种重量，输出不能称出的总数和类别*/


int main()
{
	int n;
	while( scanf("%d", &n)!=EOF )
	{
		memset(num, 0, sizeof(num) );
		memset(c1, 0, sizeof(c1) );
		memset(c2, 0, sizeof(c2) );

		int total = 0;
		for(int i=0; i<n; i++)							//重量输入
		{scanf("%d", &num[i]);	total+=num[i];}

		//for(int i=0; i<=total; i++)
		//c1[i] = c2[i] = 0;
		c1[0] = 1;										//第一个括号，重量为0的
		
		for(int i=0; i<n; i++)
		{
			for(int j=0; j<=total; j++)
			{
				c2[j] |= c1[j];							//否能称得，自己砝码的重量
				if(num[i] + j <=total)					//重量和是否大于总重量
					c2[j+num[i]] |= c1[j];				//重量和
						c2[abs(j-num[i])] |= c1[j];		//重量差
			}
			for(int j=0; j<=total; j++)					//一次循环赋值
			{c1[j] = c2[j];		c2[j] = 0;}
		}
		int count = 0;
		for(int i=0; i<=total; i++)						//查找能否称得的重量
		{
			if(!c1[i])
				c2[count++] = i;
		}
		printf("%d\n", count);
		for(int i=0; i<count-1; i++)
			printf("%d ", c2[i]);
		if(count) printf("%d\n", c2[count-1]);

	}
	return 0;
}


/*
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
*/

来自为知笔记(Wiz)

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