During the research on properties of the greatest common divisor (GCD) of a set of numbers, Ildar, a famous mathematician, introduced a brand new concept of the weakened common divisor (WCD) of a list of pairs of integers.
For a given list of pairs of integers (a1,b1)(a1,b1), (a2,b2)(a2,b2), ..., (an,bn)(an,bn) their WCD is arbitrary integer greater than 11, such that it divides at least one element in each pair. WCD may not exist for some lists.
For example, if the list looks like [(12,15),(25,18),(10,24)][(12,15),(25,18),(10,24)], then their WCD can be equal to 22, 33, 55 or 66 (each of these numbers is strictly greater than 11 and divides at least one number in each pair).
You're currently pursuing your PhD degree under Ildar's mentorship, and that's why this problem was delegated to you. Your task is to calculate WCD efficiently.
The first line contains a single integer nn (1≤n≤1500001≤n≤150000) — the number of pairs.
Each of the next nn lines contains two integer values aiai, bibi (2≤ai,bi≤2⋅1092≤ai,bi≤2⋅109).
Print a single integer — the WCD of the set of pairs.
If there are multiple possible answers, output any; if there is no answer, print −1−1.
3
17 18
15 24
12 15
6
2
10 16
7 17
-1
5
90 108
45 105
75 40
165 175
33 30
5
In the first example the answer is 66 since it divides 1818 from the first pair, 2424 from the second and 1212 from the third ones. Note that other valid answers will also be accepted.
In the second example there are no integers greater than 11 satisfying the conditions.
In the third example one of the possible answers is 55. Note that, for example, 1515 is also allowed, but it's not necessary to maximize the output.
http://codeforces.com/contest/1025/problem/B
大意：求出weakened common divisor (WCD)
给出n对数，他们的WCD为可以将每一对数中至少一个数整除的数（1除外），如果不存在则输出-1.存在多个则输出任意一个。
两种方法：
1.将第一对数保留（a，b）,后面每一对数相乘，变成一个数x。然后更新（a，b）,将其替换成gcd(a,x)，gcd(b,x)。最后再输出a或者b的最小因子
代码：
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll a,b,n;
ll gcd(ll x,ll y){return x%y? gcd(y,x%y):y;}
int main(){
cin>>n;
cin>>a>>b;
for(int i=1;i<n;++i){
ll x,y;cin>>x>>y;
x*=y;a=gcd(a,x);b=gcd(b,x);
}
if(a==1&&b==1){cout<<"-1\n";return 0;}
for(ll i=2;i*i<=a||i*i<=b;++i)
if(a%i==0||b%i==0){cout<<i<<endl;return 0;}
if(a!=1)cout<<a<<endl;
else cout<<b<<endl;
return 0;
}
方法2.
把第一对数的所有因子提出来然后对后面的每一对数进行测试，如果后面每一对数中只是有一个可以被因子可以整除，则输出因子。
代码：
#include<bits/stdc++.h>
int n,a[150010],b[150010],p[100],c;
void d(int x){
for(int i=2;1ll*i*i<=x;i++)if(x%i==0){
p[c++]=i;
while(x%i==0)x/=i;
}
if(x>1)p[c++]=x;
}
int main(){
scanf("%d",&n);
for(int i=0;i<n;i++)scanf("%d%d",a+i,b+i);
d(a[0]);
d(b[0]);
for(int i=0;i<c;i++){
bool fl=1;
for(int j=0;j<n&&fl;j++)if(a[j]%p[i]!=0&&b[j]%p[i]!=0)fl=0;
if(fl){
printf("%d\n",p[i]);
return 0;
}
}
puts("-1");
}
C. Plasticine zebra
http://codeforces.com/contest/1025/problem/C
题目大意：输出一个字符串中最长“斑马条纹”的长度，“斑马条纹”指类似于“wbwbwbw”（7）、“bwbwb”（5）。
同时为了最终获得更长的长度，可以在组装斑马之前，Grisha可以进行以下操作0或更多次。他将序列在某个地方分成两部分，然后将它们中的每一部分反转并再次将它们粘在一起。例如，如果Grisha的顺序为“ bwbbw ”（这里' b '表示黑条，' w '表示白条），那么他可以将序列拆分为bw | bbw（此处垂直条代表切割），反转两个部分并获得“ wbwbb ”。
例如
input:
bwwwbwwbw
output:
5
备注：在第一个例子中，可能的操作顺序之一是bwwbww | bw → w | wbwwwbwb → wbwbw wwbw，给出的答案等于5。
思路：
从样例一中逆推，发现最后得到的斑马条纹其实始终来自序列的两侧，经过操作后拼接在一起，于是我们可以直接将两个s拼接在一起，创造出一个2s，此时找出2s中的最长斑马条纹即可。
但要注意最长长度不会超过s的长度，就错在这里没有过fst。
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll mod=1e9+7;
const int maxn=1e7+50;
const ll inf=0x3f3f3f3f3f3f;
int k;
ull gcd(ull a,ull b)
{
while(b)
{
int t=a%b;
a=b;
b=t;
}
return a;
}
ull lcm(ull a,ull b){
return a/gcd(a,b)*b;
}
ull sum[maxn];
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(0);
std::cout.tie(0);
string s;
int num=1,maxnum=0;
cin>>s;
s=s+s;
for(int i=1;i<s.size();i++)
{
if(s[i]!=s[i-1])
{
num++;
}
else{
num=1;
}
if(num>maxnum)
{
maxnum=num;
}
}
int k=s.size();
cout<<min(maxnum,k/2)<<endl;
return 0;
}