There's no end to revenge.
Pigs like to eat birds' eggs, so birds decide to revenge.
For this, pigs decide to revenge as well.
Pigs give a task to a birdlike pig. They let he sneak into the birds group so that they can destroy the birds group.
Every bird has a ID number (It may not be unique). To let pigs know, the birdlike pig makes the "AND operation" amount all the birds' ID number and sign himself to the answer.
For example:
Birds are signed as 5, 3, 4. So the birdlike pig signs himself as (5 & 3 & 4 = 0).
One day, a group birds pass by pigs' home. So the angry pig want to know whether the birdlike pig is in.
輸入
This problem has several cases. The first line of each case is an integer N (2 <= N <= 100 000).
Then follows a line with N integers, indicates the ID of each bird/birdlike pig.
輸出
For each case, if you can find the birdlike pig then output the ID of birdlike pig. Or output 'CAUTION: NO BIRDLIKE'.
樣例輸入
5
4 0 3 4 5
樣例輸出
0
提示
無來源
XadillaX
題意 ： 輸入n 之後n個數 其中可能存在一個數是另外n-1個數的and 即&操作的結果 如果存在一個這樣數 輸出他 否則輸出CAUTION: NO BIRDLIKE 思路：對於第i個數 只要驗證 其前面的所有數的and 和後面所有數的and的結果進行and操作後 看結果是否等於第i個數是則輸出他 我們求出 1->n的連續的數的and 存到數組a裡 求出n->1連續的數的and 存進b數組 按照上面思路找即可 詳細 看代碼 代碼一看就明白了
[cpp]
#include<stdio.h>
int a[211111],b[211111],c[211111]; int main()
{ int n,num,i,flag;
while(scanf("%d",&n)!=EOF)
{ flag=0;
scanf("%d",&c[1]);
a[1]=c[1];
for(i=2;i<=n;i++)
{ scanf("%d",&c[i]);
a[i]=a[i-1]&c[i]; }
b[n]=c[n];
for(i=n-1;i>=1;i--)
{ b[i]=b[i+1]&c[i]; }
for(i=2;i<=n-1;i++)
{ if(c[i]==(a[i-1]&b[i+1]))//注意a[i-1]&b[i+1] 一定要用括號括起來 否則運算順序就變了
{ printf("%d\n",c[i]);
flag=1;
break; } }
if(flag==1) continue;
if(c[n]==a[n-1])
{ printf("%d\n",c[n]);
flag=1;
continue; }
if(flag==1) continue;
if(c[1]==b[2]) {
printf("%d\n",c[1]);
flag=1; continue;
} printf("CAUTION: NO BIRDLIKE\n");
} return 0; } #include<stdio.h>
int a[211111],b[211111],c[211111];
int main()
{
int n,num,i,flag;
while(scanf("%d",&n)!=EOF)
{
flag=0;
scanf("%d",&c[1]);
a[1]=c[1];
for(i=2;i<=n;i++)
{
scanf("%d",&c[i]);
a[i]=a[i-1]&c[i];
}
b[n]=c[n];
for(i=n-1;i>=1;i--)
{
b[i]=b[i+1]&c[i];
}
for(i=2;i<=n-1;i++)
{
if(c[i]==(a[i-1]&b[i+1]))//注意a[i-1]&b[i+1] 一定要用括號括起來 否則運算順序就變了
{
printf("%d\n",c[i]);
flag=1;
break;
}
}
if(flag==1) continue;
if(c[n]==a[n-1])
{
printf("%d\n",c[n]);
flag=1;
continue;
}
if(flag==1) continue;
if(c[1]==b[2])
{
printf("%d\n",c[1]);
flag=1;
continue;
}
printf("CAUTION: NO BIRDLIKE\n");
}
return 0;
