2014
03-09

# List Operations

A list is a sequence of zero or more elements, expressed in this form: [a1, a2, a3, ... , an], where each ai is one or more consecutive digits or lowercase letters. i.e. each list begins with a left square bracket, then zero or more elements separated by a single comma, followed by a right square bracket. There will be no whitespace characters (spaces, TABs etc) within a list.

In this problem, we use two list operations: append (++) and remove (–).
1. A ++ B: append elements in B to the end of A.
2. A — B: remove all the elements in B, from A. If something appears more than once in B, remove it that many times in A. If there are many equal elements in A to choose from, remove them from left to right (until all occurrences are removed, or there is no need to remove more elements).

Your task is to write a calculator, evaluating simple expressions or the form “list1 ++ list2” or “list1 — list2”.

There will be at most 10 expressions, one in each line, each having the form “list1 ++ list2” or “list1 — list2”, with no more than 80 characters in total (not counting the newline character). There will be exactly two spaces in each line: one before and one after the operator. Input ends with a single dot. The input is guaranteed to satisfy the restrictions stated above

There will be at most 10 expressions, one in each line, each having the form “list1 ++ list2” or “list1 — list2”, with no more than 80 characters in total (not counting the newline character). There will be exactly two spaces in each line: one before and one after the operator. Input ends with a single dot. The input is guaranteed to satisfy the restrictions stated above

[1,2,3] ++ [1,2,3]
[a,b,c,t,d,e,t,x,y,t] -- [t]
[a,b,c,t,d,e,t,x,y,t] -- [t,t,t,t]
[123] ++ [456]
.

[1,2,3,1,2,3]
[a,b,c,d,e,t,x,y,t]
[a,b,c,d,e,x,y]
[123,456]
4

->题目请戳这里<-

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int N = 10005;
char str[N];
struct node
{
char s[81];
int len;
int flag;
}list1[N],list2[N];
int n1,n2;
int op;

void solve()
{
int len = strlen(str);
memset(list1,0,sizeof(list1));
memset(list2,0,sizeof(list2));
n1 = n2 = 0;
int i,j;
for(i = 1;i < len;)
{
while(str[i] != ',' && str[i] != ']')
list1[n1].s[list1[n1].len ++] = str[i ++];
list1[n1].s[list1[n1].len] = '\0';
n1 ++;
if(str[i] == ']')
break;
i ++;
}
i +=2;
if(str[i] == '+')
op = 1;
else
op = 0;
i += 4;
for(;i < len - 1;)
{
while(str[i] != ',' && str[i] != ']')
list2[n2].s[list2[n2].len ++] = str[i ++];
list2[n2].s[list2[n2].len] = '\0';
n2 ++;
i ++;
}
if(op == 1)
{
printf("[%s",list1[0].s);
for(i = 1;i < n1;i ++)
printf(",%s",list1[i].s);
for(i = 0;i < n2;i ++)
printf(",%s",list2[i].s);
printf("]\n");
return;
}
for(i = 0;i < n2;i ++)
{
for(j = 0;j < n1;j ++)
{
if(list1[j].flag)
continue;
if(strcmp(list1[j].s,list2[i].s) == 0)
{
list1[j].flag = 1;
break;
}
}
}
printf("[");
for(j = n1 - 1;j >= 0;j --)
if(list1[j].flag == 0)
break;
for(i = 0;i < j;i ++)
{
if(list1[i].flag == 0)
printf("%s,",list1[i].s);
}
printf("%s]\n",list1[j].s);
}

int main()
{
while(gets(str))
{
if(str[0] == '.')
break;
solve();
}
return 0;
}
/*
[1,2,3,2,3] -- [2,3,2]
*/

1. 问题的关键就是这个世界上不只你一个人 如果真的每个人都向往和平 。。。但是会吗 有的人跟你志同道合 有的却偏要跟你针锋相对！ 本来就是内心造成的，本来就是人心的问题，本来就是这个问题的根源解决不了 这一切都不是废话吗？ 平常我们总是说某人心善 某人心恶

2. 歪果仁是觉得过26岁就完了，因为不属于“年轻人”范畴了：坐车不能买半票，旅行不能住青年旅社，银行存款也不能税务减免了

3. 歪果仁是觉得过26岁就完了，因为不属于“年轻人”范畴了：坐车不能买半票，旅行不能住青年旅社，银行存款也不能税务减免了

4. 歪果仁是觉得过26岁就完了，因为不属于“年轻人”范畴了：坐车不能买半票，旅行不能住青年旅社，银行存款也不能税务减免了

5. 歪果仁是觉得过26岁就完了，因为不属于“年轻人”范畴了：坐车不能买半票，旅行不能住青年旅社，银行存款也不能税务减免了

6. 歪果仁是觉得过26岁就完了，因为不属于“年轻人”范畴了：坐车不能买半票，旅行不能住青年旅社，银行存款也不能税务减免了

7. 歪果仁是觉得过26岁就完了，因为不属于“年轻人”范畴了：坐车不能买半票，旅行不能住青年旅社，银行存款也不能税务减免了

8. 站长，你好！
你创办的的网站非常好，为我们学习算法练习编程提供了一个很好的平台，我想给你提个小建议，就是要能把每道题目的难度标出来就好了，这样我们学习起来会有一个循序渐进的过程！

9. 漂亮。佩服。
P.S. unsigned 应该去掉。换行符是n 不是/n
还可以稍微优化一下，
int main() {
int m,n,ai,aj,bi,bj,ak,bk;
while (scanf("%d%d",&m,&n)!=EOF) {
ai = sqrt(m-1);
bi = sqrt(n-1);
aj = (m-ai*ai-1)>>1;
bj = (n-bi*bi-1)>>1;
ak = ((ai+1)*(ai+1)-m)>>1;
bk = ((bi+1)*(bi+1)-n)>>1;
printf("%dn",abs(ai-bi)+abs(aj-bj)+abs(ak-bk));
}
}

10. 算法是程序的灵魂，算法分简单和复杂，如果不搞大数据类，程序员了解一下简单点的算法也是可以的，但是会算法的一定要会编程才行，程序员不一定要会算法，利于自己项目需要的可以简单了解。