2014
03-16

# Calculate the expression

You may find it’s easy to calculate the expression such as:
a = 3
b = 4
c = 5
a + b + c = ?
Isn’t it?

The first line contains an integer stands for the number of test cases.
Each test case start with an integer n stands for n expressions will follow for this case.
Then n � 1 expressions in the format: [variable name][space][=][space][integer] will follow.
You may suppose the variable name will only contain lowercase letters and the length will not exceed 20, and the integer will between -65536 and 65536.
The last line will contain the expression you need to work out.
In the format: [variable name| integer][space][+|-][space][variable name| integer] …= ?
You may suppose the variable name must have been defined in the n � 1 expression and the integer is also between -65536 and 65536.

The first line contains an integer stands for the number of test cases.
Each test case start with an integer n stands for n expressions will follow for this case.
Then n � 1 expressions in the format: [variable name][space][=][space][integer] will follow.
You may suppose the variable name will only contain lowercase letters and the length will not exceed 20, and the integer will between -65536 and 65536.
The last line will contain the expression you need to work out.
In the format: [variable name| integer][space][+|-][space][variable name| integer] …= ?
You may suppose the variable name must have been defined in the n � 1 expression and the integer is also between -65536 and 65536.

3
4
aa = 1
bb = -1
aa = 2
aa + bb + 11 = ?
1
1 + 1 = ?
1
1 + -1 = ?

12
2
0

#include"stdio.h"
#include"string.h"
struct node
{
char str[21];
int x;
}A[1001];
int main()
{
int t,n;
int i,j;
int cnt,a;
char s[22];
scanf("%d",&t);
while(t--)
{
cnt=0;
scanf("%d",&n);
getchar();
for(i=0;i<n-1;i++)
{
scanf("%s = %d",s,&a);
getchar();
for(j=0;j<cnt;j++)
{
if(strcmp(A[j].str,s)==0)
{
A[j].x=a;break;
}
}
if(j==cnt)
{
strcpy(A[cnt].str,s);
A[cnt].x=a;
cnt++;
}
}
int ans,f;
f=ans=0;
while(1)
{
scanf("%s",s);
if(strcmp("?",s)==0)break;
if(strcmp("=",s)==0)printf("%d\n",ans);
else if(strcmp("+",s)==0)f=1;
else if(strcmp("-",s)==0)f=-1;
else
{
for(i=0;i<cnt;i++)
if(strcmp(A[i].str,s)==0)
break;
if(i==cnt)
{
int t,ff;
t=0;ff=1;
for(i=0;s[i];i++)
{
if(s[i]=='-'){ff=-1;continue;}
t=t*10+s[i]-'0';
}
strcpy(A[cnt].str,s);
A[cnt].x=t;
A[cnt].x*=ff;
cnt++;
i=cnt-1;
}
if(f==0)ans=A[i].x;
else if(f==1)ans=ans+A[i].x;
else if(f==-1)ans=ans-A[i].x;
}
}
}
return 0;
}

1. 我没看懂题目
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
我觉得第一个应该是5 6 -1 5 4 输出是19 5 4
第二个是7 0 6 -1 1 -6 7输出是14 7 7
不知道题目例子是怎么得出来的

2. 老实说，这种方法就是穷举，复杂度是2^n，之所以能够AC是应为题目的测试数据有问题，要么数据量很小，要么能够得到k == t，否则即使n = 30，也要很久才能得出结果，本人亲测