题目描述
A string is perfect if it has the smallest lexicographical ordering among its cyclic rotations.
For example: "0101" is perfect as it is the smallest string among ("0101", "1010", "0101", "1010").
Given a 01 string, you need to split it into the least parts and all parts are perfect.
输入描述:
The first line of the input gives the number of test cases, T (T≤300)T\ (T \leq 300)T (T≤300). test cases follow. For each test case, the only line contains one non-empty 01 string. The length of string is not exceed 200.
输出描述:
For each test case, output one string separated by a space.
示例1
输入
复制
4 0 0001 0010 111011110
输出
复制
0 0001 001 0 111 01111 0
给定一个字符串，将其划分成几部分，每部分都满足循环串字典序最小，如何划分使划分次数最少
枯了 暴力枚举子串终点，向后续延
#include <bits/stdc++.h>
using namespace std;
string s;
int len;
bool check(int l,int r)
{
int len=r-l+1;
string s1=s.substr(l,len);
string s2=s1+s1;
for(int i=0;i<len;i++)
{
if(s2.substr(i,len)<s1)
return false;
}
return true;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie();
int t,n;
cin>>t;
while(t--)
{
cin>>s;
len=s.size();
int l=0,r;
for(;l<len;l++)
{
r=len-1;
while(!check(l,r))
r--;
string tmp=s.substr(l,r-l+1);
cout<<tmp;
if(r!=len-1)
cout<<' ';
l=r;
}
cout<<'\n';
}
return 0;
}
题目描述
In mathematics, a polynomial is an expression consisting of variables (also called indeterminate) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. For example, x2+4x+7x^2 + 4x + 7x2+4x+7.
A polynomial is said to be irreducible if it cannot be factored into two or more non-trivial polynomials with real coefficients.
For example, x2+1x^2+1x2+1 is irreducible, but x2−2x+1x^2-2x+1x2−2x+1 is not (since x2−2x+1=(x−1)(x−1)x^2-2x+1=(x-1)(x-1)x2−2x+1=(x−1)(x−1)).
Given a polynomial of degree with integer coefficients: anxn+an−1xn−1+...+a1x+a0a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0anxn+an−1xn−1+...+a1x+a0, you need to check whether it is irreducible or not.
输入描述:
The first line of the input gives the number of test cases, T (T≤100)T\ (T \leq 100)T (T≤100). test cases follow.
For each test case, the first line contains one integers n (0≤n≤20)n\ (0 \leq n \leq 20)n (0≤n≤20). Next line contains n + 1n\ +\ 1n + 1 integer numbers:
an,an−1,...,a1,a0a_n, a_{n-1}, ..., a_1, a_0an,an−1,...,a1,a0
−109≤ai≤109-10^9 \leq a_i \leq 10^9−109≤ai≤109
an≠0a_n \ne 0an=0 输出描述:
For each test case, output "Yes" if it is irreducible, otherwise "No".
示例1
输入
复制
2 2 1 -2 1 2 1 0 1
输出
复制
No Yes
询问一个多项式是否可拆分成单项式乘积的形式
我好菜啊太菜了
实数域不可拆分多项式只有两种：一次多项式和二次的(b^2<4ac)
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t,n,a[25];
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=1;i<=n+1;i++)
scanf("%d",&a[i]);
if(n>2)
cout<<"No"<<'\n';
else if(n<2)
cout<<"Yes"<<'\n';
else
{
if(a[2]*a[2]-4*a[1]*a[3]>=0)
cout<<"No"<<'\n';
else
cout<<"Yes"<<'\n';
}
}
return 0;
}
题目描述
I have a very simple problem for you. Given a positive integeter n (1≤n≤1000000)n\ (1 \leq n \leq 1000000)n (1≤n≤1000000) and a prime number p (2≤p≤1000000)p\ (2 \leq p \leq 1000000)p (2≤p≤1000000), your job is to output a positive number which is divisible by and has exactly
digits. Please output "T_T" if you can not find such number.
输入描述:
The first line of the input file contains two integers n (1≤n≤1000000)n\ (1 \leq n \leq 1000000)n (1≤n≤1000000) and p (2≤p≤1000000)p\ (2 \leq p \leq 1000000)p (2≤p≤1000000). is a prime number.
输出描述:
Output one number (without leading zeros) or "T_T"
示例1
输入
复制
2 5
输出
复制
10
示例2
输入
复制
1 11
输出
复制
T_T
示例3
输入
复制
5 2
输出
复制
10000
给定一个素数p，输出一个n位的p的倍数
#include <bits/stdc++.h>
using namespace std;
int main()
{
int p,n;
while(scanf("%d%d",&n,&p)!=EOF)
{
int cnt=0;
int tmp=p;
while(p>0)
{
cnt++;
p/=10;
}
if(cnt>n)
cout<<"T_T"<<'\n';
else
{
cout<<tmp;
n-=cnt;
while(n--)
cout<<0;
cout<<'\n';
}
}
return 0;
}
题目描述
Reversed number is a number written in arabic numerals but the order of digits is reversed. The first digit becomes last and vice versa. And all the leading zeros are omitted.
For example: the reversed number of 1234 is 4321. The reversed number of 1000 is 1.
We define reversed number of as
. Given you two positive integers
and
, you need to calculate the reversed sum:
.
输入描述:
The first line of the input gives the number of test cases, T (T≤300)T\ (T \leq 300)T (T≤300). test cases follow.
For each test case, the only line contains two positive integers: and . (1≤A,B≤231−11 \leq A, B \leq 2^{31}-11≤A,B≤231−1) 输出描述:
For each test case, output a single line indicates the reversed sum.
示例1
输入
复制
3 12 1 101 9 991 1
输出
复制
22 11 2
A的反转加B的反转再反转
#include <bits/stdc++.h>
using namespace std;
int main()
{
long long a,b,ans,tmp;
int t;
scanf("%d",&t);
while(t--)
{
scanf("%lld%lld",&a,&b);
long long c=0,d=0;
while(a)
{
c*=10;
c+=a%10;
a/=10;
}
while(b)
{
d*=10;
d+=b%10;
b/=10;
}
tmp=c+d;
ans=0;
while(tmp)
{
ans*=10;
ans+=tmp%10;
tmp/=10;
}
cout<<ans<<'\n';
}
return 0;
}
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