zzuoj--10399--Turing equation（模拟）

Turing equation

Time Limit: 1 Sec Memory Limit:
128 MB

Submit: 152 Solved: 85

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Description

The fight goes on, whether to store numbers starting with their most significant digit or their least significant digit. Sometimes this is also called the "Endian War". The battleground dates far back into
the early days of computer science. Joe Stoy, in his (by the way excellent) book "Denotational Semantics", tells following story:

"The decision which way round the digits run is, of course, mathematically trivial. Indeed, one early British computer had numbers running from right to
left (because the spot on an oscilloscope tube runs from left to right, but in serial logic the least significant digits are dealt with first). Turing used to mystify audiences at public lectures when, quite by accident, he would slip into this mode even
for decimal arithmetic, and write things like 73+42=16. The next version of the machine was made more conventional simply by crossing the x-deflection wires: this, however, worried the engineers, whose waveforms were all backwards. That problem was
in turn solved by providing a little window so that the engineers (who tended to be behind the computer anyway) could view the oscilloscope screen from the back.

You will play the role of the audience and judge on the truth value of Turing's equations.

Input

The input contains several test cases. Each specifies on a single line a Turing equation. A Turing equation has the form "a+b=c", where a, b, c are numbers made up of the digits 0,...,9. Each number will consist
of at most 7 digits. This includes possible leading or trailing zeros. The equation "0+0=0" will finish the input and has to be processed, too. The equations will not contain any spaces.

Output

For each test case generate a line containing the word "TRUE" or the word "FALSE", if the equation is true or false, respectively, in Turing's interpretation, i.e. the numbers being read backwards.

Sample Input

73+42=16
5+8=13
0001000+000200=00030
0+0=0

Sample Output

TRUE
FALSE
TRUE

HINT

Source

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define  PI(x) printf("%d",x)
#define  PL(x) printf("%lld",x)
#define P_ printf(" ")
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
typedef long long LL;
char s[35],t[10];
int ans[3];
int main()
{
while(scanf("%s",s),strcmp(s,"0+0=0"))
{
int k=0,tp=0,temp=0;
for(int i=0;s[i];i++)
{
if(isdigit(s[i]))
{
t[k++]=s[i];
}
else
{
reverse(t,t+k);
for(int j=0;j<k;j++)
temp=temp*10+t[j]-'0';
ans[tp++]=temp;
k=0;temp=0;
}
}
reverse(t,t+k);
for(int j=0;j<k;j++)
temp=temp*10+t[j]-'0';
ans[tp++]=temp;
if(ans[0]+ans[1]==ans[2])
puts("TRUE");
else puts("FALSE");
}
return 0;
}