Portal Title (title 72 inside)
Input Format
$ Bit.in $ input file
for each input file contains $ T $ test cases. The first line of the input file integer $ T $, indicates the number of data sets. Then $ T $ rows, each row represents a set of test data comprising each test three numbers separated by spaces $ ResultAnd, ResultOr, ResultXor $, expressed sequence $ a \ & b, a | b, a \ text { b $} ^ limits of.
If three numbers of an item is $ -1 $, then that this is not a limitation. Otherwise the number must be a non-negative integer, A represents $ $ $ b $ and results after this operation.
Output Format
Bit.out input file $ $
$ T $ lines, each an integer or a string $ "inf" $ (output without the quotes).
Integer or string of $ I $ $ row indicates the answer to the first test data set I $.
Sample
Sample input:
10
20 1015 995
921 661 -1
-1 375 -1
30445634 30446311 30446245
11997588 11998143 555
-1 65535 -1
-1 465530605 312684161
2118209 930739953 928621744
69739040 402620388 332881348
4594346 533159678 528565332
Sample output:
128
0
2187
0
32
43046721
4096
32768
131072
4096
Data range and tips
Sample explained:
Swatch $ T = 10 $, $ I $ row and the first sample is a set of input data from the first $ (2 * i-1) $ test points
data range:
For all the test points: $ T \ leqslant 10, ResultAnd , ResultOr, ResultXor $ all $ [- 1, 10 ^ 9] integer in the range of $. On the same set of data, $ ResultAnd, ResultOr, ResultXor $ are not $ -1 $.
The first $ 1 $ and $ 5 $ test points: $ 0 \ leqslant ResultOr \ leqslant 1023 $
of $ 6 $ to $ 10 $ test points: $ ResultOr-ResultAnd \ leqslant 1000 , ResultOr> = 0, ResultAnd> = 0 $
of $ 11 $ to $ 16 $ test points: the answer does not appear $ INF $
of $ 11 $ and $ 12 $ test points further satisfies: $ ResultAnd = ResultXor = -1 $
of $ 17, $ test point satisfies: is not equal to $ INF $ of the answer will not exceed the range of $ int $ data type can represent the
first $ 19,20 $ test points: no special restrictions
answer
Vigorously Category talk to.
Bit operation must first be extracted for each treatment.
We may assume $ a [i], b [i] .c [i] $ denote $ \ &, |, \ text {^} $ each bit.
Followed by enumeration:
. $ \ Alpha $ has a limit: $ ans = 2 ^ {\ sum \ limits_ {i = 1} ^ {30} a [i] \ text {^} b [i]} $.
$\beta.\&$没有限制：$ans=2^{\sum \limits_{i=1}^{30}b[i]\&c[i]}$。
$\gamma.|$没有限制：$ans=2^{\sum \limits_{i=1}^{30}c[i]}$。
$\delta.\text{^}$没有限制：$ans=2^{\sum \limits_{i=1}^{30}a[i]\text{^}b[i]}$。
$\epsilon.$只有$\&$有限制：$ans=inf$。
$\zeta.$只有$|$有限制：$ans=3^{\sum \limits_{i=1}^{30}b[i]}$。
$eta.$只有$\text{^}$有限制：$ans=inf$。
时间复杂度：$\Theta(30\times T)$。
期望得分：$100$分。
实际得分：$100$分。
代码时刻
#include<bits/stdc++.h>
using namespace std;
int AND,OR,XOR;
int a[50],b[50],c[50];
long long qpow(long long x,long long y)
{
long long res=1;
while(y)
{
if(y&1)res=res*x;
x=x*x;
y>>=1;
}
return res;
}
int main()
{
int T;scanf("%d",&T);
while(T--)
{
int sum=0;
scanf("%d%d%d",&AND,&OR,&XOR);
if(AND!=-1&&OR==-1&&XOR==-1){puts("inf");continue;}
if(AND==-1&&OR==-1&&XOR!=-1){puts("inf");continue;}
if(AND!=-1&&OR!=-1&&XOR!=-1)
{
for(int i=0;i<=30;i++)
{
a[i]=(AND>>i)&1;
b[i]=(OR>>i)&1;
c[i]=(XOR>>i)&1;
if(a[i]&&!b[i]){puts("0");goto nxt;}
if(a[i]&&c[i]){puts("0");goto nxt;}
if(!b[i]&&c[i]){puts("0");goto nxt;}
sum+=a[i]^b[i];
}
printf("%lld\n",qpow(2,sum));
}
if(AND!=-1&&OR==-1&&XOR!=-1)
{
for(int i=0;i<=30;i++)
{
a[i]=(AND>>i)&1;
c[i]=(XOR>>i)&1;
if(a[i]&&c[i]){puts("0");goto nxt;}
sum+=c[i];
}
printf("%lld\n",qpow(2,sum));
}
if(AND==-1&&OR!=-1&&XOR!=-1)
{
for(int i=0;i<=30;i++)
{
b[i]=(OR>>i)&1;
c[i]=(XOR>>i)&1;
if(!b[i]&&c[i]){puts("0");goto nxt;}
sum+=b[i]&c[i];
}
printf("%lld\n",qpow(2,sum));
}
if(AND!=-1&&OR!=-1&&XOR==-1)
{
for(int i=0;i<=30;i++)
{
a[i]=(AND>>i)&1;
b[i]=(OR>>i)&1;
sum+=a[i]^b[i];
if(a[i]&&!b[i]){puts("0");goto nxt;}
}
printf("%lld\n",qpow(2,sum));
}
if(AND==-1&&OR!=-1&&XOR==-1)
{
for(int i=0;i<=30;i++)
sum+=(OR>>i)&1;
printf("%lld\n",qpow(3,sum));
}
nxt:;
}
return 0;
}
rp++