1. Exclusive square number
Backtracking + pruning
#include <iostream>
#include <vector>
using namespace std;
const int N=10;
bool st[N];
vector<vector<int>>ans;
vector<int>num;
void backtracking(int level){
if(level==6){
long long tmp=0;
bool flag=true;
for(int i=0;i<6;i++){
tmp*=10;
tmp+=num[i];
}
tmp*=tmp;
while(tmp){
if(st[tmp%10]){
flag=false;
break;
}
tmp/=10;
}
if(flag)ans.push_back(num);
return;
}
for(int i=0;i<=9;i++){
if(!st[i]){
st[i]=true;
num.push_back(i);
if(!(level==0&&i==0))backtracking(level+1);//排除最高位为0
num.pop_back();
st[i]=false;
}
}
}
int main()
{
ans.clear();
num.clear();
backtracking(0);
int s=0;
for(int i=0;i<ans.size();i++){
for(int j=0;j<ans[i].size();j++){
s*=10;
s+=ans[i][j];
}
if(s!=203879)break;
else s=0;
}
printf("%d",s);
return 0;
}
2. The quantity that cannot be bought
Method 1: Do your best to analyze.
When the greatest common divisor d=(p,q)>1, there is no solution. All coprime numbers must have solutions
. eg. 2 4 -> All numbers that are not multiples of 2 cannot be made up.
Method 2: Play the watch to find the rule
, play the watch code
#include <iostream>
using namespace std;
bool dfs(int i,int p,int q){
if(i==0)return true;
if(i>=p&&dfs(i-p,p,q))return true;
if(i>=q&&dfs(i-q,p,q))return true;
return false;
}
int main()
{
int p,q;
scanf("%d%d",&p,&q);
int res=0;
for(int i=0;i<=1000;i++){
if(!dfs(i,p,q))res=i;
}
printf("%d",res);
return 0;
}
Pei's theorem
(p, q) = d
, then there must be 2 integers such that ap+bq=d
If (p, q)=q
must exist ap+bq=1 ,
we must make up m
as long as there is amp+bmp=m
(am- q)p+(bm+q)p=m
Finally i=(p-1)(q-1)-1
#include <iostream>
using namespace std;
int main()
{
int p,q;
scanf("%d%d",&p,&q);
printf("%d",(p-1)*(q-1)-1);
return 0;
}
3. Palindrome date
#include <iostream>
#include <vector>
using namespace std;
const int N=99999999;
int month[13]={
0,31,28,31,30,31,30,31,31,30,31,30,31};
//判断日期是否合法
bool checkIsDate(int num){
int yyyy=num/10000;
int mm=num/100%100;
if(mm>12||mm==00)return false;
int dd=num%100;
if(mm==2){
if(yyyy%400==0||(yyyy%4==0&&yyyy%100!=0))month[2]=29;
else month[2]=28;
}
if(dd>month[mm]||dd<=0)return false;
return true;
}
//判断是否是回文数
bool checkIsH(int num){
vector<int>ans;
for(int i=0;i<8;i++){
ans.push_back(num%10);
num/=10;
}
for(int i=0;i<4;i++){
if(ans[i]!=ans[7-i])return false;
}
return true;
}
//判断是否是ABABBABA
bool checkIsAB(int num){
vector<int>ans;
for(int i=0;i<8;i++){
ans.push_back(num%10);
num/=10;
}
if(ans[0]==ans[1])return false;
if(ans[0]!=ans[2]||ans[2]!=ans[5]||ans[5]!=ans[7])return false;
if(ans[1]!=ans[3]||ans[3]!=ans[4]||ans[4]!=ans[6])return false;
return true;
}
int main()
{
int start;
scanf("%d",&start);
int ans1=0,ans2=0;
bool flag1=true,flag2=true;
for(int i=start+1;i<=N;i++){
if(checkIsDate(i)){
if(checkIsH(i)){
if(flag1){
ans1=i;
flag1=false;
}
if(checkIsAB(i)){
if(flag2){
ans2=i;
flag2=false;
break;
}
}
}
}
}
printf("%d\n%d",ans1,ans2);
return 0;
}
4. Joseph Ring
Refer to the detailed explanation of the Joseph ring problem.
When it is not difficult to deduce the last remaining element, the serial number is 0
, that is, J2(1) = 0------(1)
Then we know that the new queue obtained in this way is also easy to know How to push back:
On the other hand, the above changes are all minus a q . So:
push the last remaining 0 before the serial number:
the formula to change back is as follows: old = (new + q) % n
When there are two remaining elements (ie 0 ,1), delete the sequence number 0, and then change the sequence number after 0 to 0 (ie 1 becomes 0)
J2(2) = (J2(1) + 2) % 2 = 0------- ----(2)
When there are three remaining elements (0,1,2), delete the sequence number 2, and then change the sequence number after 2 to 0 (that is, 0 becomes 0)
J2(3) = (J2(2) + 2) % 3 = 2
When there are two remaining elements (0, 1, 2, 3), delete the sequence number of 0, and then change the sequence number after 0 to 0 (that is, 1 becomes 0)
J2(4) = (J2(3) + 2) % 4 = 0
```cpp
#include<iostream>
#include<stdio.h>
using namespace std;
int f(int n,int m){
if(n == 1){
return 0; //这里返回下标,从0开始，只有一个元素就是剩余的元素0
}
else{
return (f(n-1,m) + m) % n; //我们传入的n是总共多少个数
}
}
int main(){
int a,b;
cin>>a>>b;
cout<<f(a,b)+1<<endl;
//或者，直接循环迭代，求出来的result如上
// int result = 0;
// for(int i = 2;i <= a;i++){
// result = (result+b) %i;
// }
return 0;
}
```