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Description
Rami went back from school and he had an easy homework about bitwise operations (and,or,..) The homework was like this : You have an equation : " A | B = C " ( this bar ‘|‘ means OR ) you are given the value of A and B , you need to find the value of C ?? as we said before this is an easy question for Rami to solve ,but he wonderd if the equation was like this: " A | C = B " and he is given the value of A and B , how many value of C exists ?? Rami wants your help in this hard question ... He will help you as much as he can ,so he will convert the two decimal numbers(A and B) to binary representaion ( like this: if A=6 –> A=110 ) and this what he knows about OR operation and might be helpfull with your task :
Input
The input consists of several test cases. The first line of the input contains a single integer T, the number of the test cases. each test case consists of 3 lines : the 1st line is the length of the binary representaion of the 2 numbers.1<=n<=100 the 2nd line is the number A in base 2 .(a string consits of 0s and 1s only ) the 3rd line is the number B in base 2 .(a string consits of 0s and 1s only )
Output
for each test case,you need to print the number of possible values of C that make the eqaution correct in a seperate line .(read page 2 carefully)
Sample Input
3
2
10
11
3
110
110
4
1110
1011
2
4
IMPOSSIBLE
Hint
as the answer may be very very large , you should print the answer mod 1000000007. there might be no answer for the equation also , in this case you should print "IMPOSSIBLE" (without qoutes).
写的时候找了半天错才发现要对求的数取模.....感觉智商被掏空
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long LL;
int main()
{
//freopen("input.txt","r",stdin);
int t;
cin>>t;
while(t--)
{
int n;
char a[105],b[105];
cin>>n;
LL res=1;
for(int i=1;i<=n;i++)
cin>>a[i];
for(int i=1;i<=n;i++)
cin>>b[i];
for(int i=1;i<=n;i++)
{
if(a[i]==‘1‘)
{
if(b[i]==‘0‘)
{
res=0;
break;
}
if(b[i]==‘1‘)
{
res*=2;
res=res%1000000007;
}
}
}
if(res)
cout<<res%1000000007<<endl;
else
cout<<"IMPOSSIBLE"<<endl;
}
}
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原文地址：http://www.cnblogs.com/Hyouka/p/5719708.html
