2014
01-04

# Balance

You are given a set of weight values and asked to measure something with a certain weight. Weight x is measurable with a set y of weight values if and only if you can do the following: Place one weight with value x on either the left or right side of the scale. Then place some combination of weights on the scale with values from the set y so that the left and right sides have the same total value. You may use each value in y zero or more times to achieve this. In order to reach the target, you choose a kind of weight randomly and try to measure with it. If you find it impossible, you choose an additional kind and try again until you can measure with the set of weights you have chosen. Now, please tell me the expect kinds of weight you have to choose.

For each case, the first line is two integers n and m (1<=n<=15), which n represents the number of weights you have and m is the weight you have to measure. Followed by n positive integers ai (1<=i<=n), indicating the mass of each kind of weight. 0 < m,ai < 2^31.

For each case, the first line is two integers n and m (1<=n<=15), which n represents the number of weights you have and m is the weight you have to measure. Followed by n positive integers ai (1<=i<=n), indicating the mass of each kind of weight. 0 < m,ai < 2^31.

1 2
3
3 3
1 2 3
3 5
2 3 4

-1
1.333
2.333

#include<stdio.h>
#include<string.h>
#include<ctype.h>
char s[1000];
int main()
{
int n;
char *p,*_p;
scanf("%d",&n);
getchar();
while(n-->0)
{
gets(s);
p=s;
while(*p!='\0')
{
if(*p=='('&&*(p+1)==')'||(*p=='['&&*(p+1)==']'))
{
*p='\0';
_p=p+2;
p=s;
strcat(p,_p);
}
else
p++;
}
if(s[0]=='\0')
printf("Yes\n");
else
printf("No\n");
}
return 0;
}

1. a是根先忽略掉，递归子树。剩下前缀bejkcfghid和后缀jkebfghicd，分拆的原则的是每个子树前缀和后缀的节点个数是一样的，根节点出现在前缀的第一个，后缀的最后一个。根节点b出现后缀的第四个位置，则第一部分为四个节点，前缀bejk，后缀jkeb，剩下的c出现在后缀的倒数第2个，就划分为cfghi和 fghic，第3部分就为c、c

2. L（X [0 .. M-1]，Y [0 .. N-1]）= 1 + L（X [0 .. M-2]，Y [0 .. N-1]）这个地方也也有笔误
应改为L（X [0 .. M-1]，Y [0 .. N-1]）= 1 + L（X [0 .. M-2]，Y [0 .. N-2]）