Title description
Given a B array B[0...n-1]. You need to find another array A[0...n-1] so that the A array meets the following conditions at the same time:
1. For 0<=i<n, A[i] must be 2^j, where j is a positive integer. That is, A[i] must be 2, 4, 8, 16, 32, 64... etc.
2. For 0<=i<n, A[i]^B[i] must divide P evenly, where P=A[0] * A[1] * A[2] * …A[n-1].
Note: For this question, all means power, that is, how many powers, for example: 2 3 means 2 to the 3rd power, so 2^3=8.
If you can find the A array that meets the above requirements, then output "Possible" or output "Impossible" without outputting double quotes.
enter
Multiple sets of test data.
In the first line, an integer G indicates that there are G groups of test data. 1 <= G <= 13.
Each test data format: the
first line, an integer n. 1 <= n <= 50.
In the second line, there are n integers, and the i-th integer is B[i]. 1 <= B[i] <= 10.
Output
There are G lines, one string per line, "Possible" or "Impossible", double quotes are not output.
Sample input Copy
4
2
3 2
3
3 3 3
2
1 10
3
2 3 10
Sample output Copy
Possible
Possible
Impossible
Possible
prompt
The first set of test data: A[] = {2,2} The
second set of test data: A[] = {2,2,2} The
fourth set of test data: A[] = {8,4,2}
Problem analysis:
We assume that the initialization state of array A is all 2, because array B has n numbers, so the initialization state of array A is n 2, and
we take the last set of test examples to derive the change process of array A. The
above is a solution. ideas
but when no solution that is output impossible
I can not program their own end, and then I added a limit, when p is greater than the index number on the impossible
With AC code
在这里插入代#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int b[1000],a[1000];
int main() {
int G;
cin >> G;
while(G--) {
int n,i,j,num;
cin>>n;
num=n;
for(i=1; i<=n; i++) {
cin>>b[i];
a[i]=b[i];
}
int temp=0,flag=1;
for(i=1; i<=n; i++) {
if(b[i]<num) {
b[i]+=a[i]; //因为没有数组A ，只能把数组A 的变化量体现在数组B上
i=0; //只要不满足条件 就从头开始重新判断
num++; // p的幂加一
temp=0;
if(num>6666) {
// P的幂加一 ，设定上限666
goto loop;
break;
}
} else
temp++; //每有一个 满足情况的数。计数器加一
}
if(temp==n) {
//当计数器的值与数组中个数相等时说明所有的数都符合条件，
printf("Possible\n"); // 即找到了符合条件的数组A
flag=0;
goto loop;
}
else
printf("Impossible\n");
loop: //添加一个标记，幂超出范围时终止程序
if(flag)
printf("Impossible\n");
}
return 0;
}
码片
Is there a better way for the big guys