#include<iomanip>
//========================
using namespace std;
class Date{
int year,month,day;
public:
bool isleapyear(){
return(year%4==0&&year%100!=0)||(year%400==0);
}
void print(){
cout<<setfill('0');
cout<<setw(4)<<year<<'-'<<setw(2)<<month<<'-'<<setw(2)<<day<<'\n';
cout<<setfill(' ');
}
void set(int y,int m,int d){
year=y; month=m; day=d;
}
};
//--------------------------
int main(){
Date kk;
int n;cin>>n;
for(int i=1;i<=n;i++){
int y;int m;int d;
cin>>y>>m>>d; // kk.set(1000,8,16);
kk.set(y,m,d);
if(kk.isleapyear()) kk.print();
else cout<<"notleap\n";
}
cout<<"over"<<'\n';
}
//--------------------------
鎯崇敤綾誨仛8涓帓搴忕畻娉?鏈夌偣娌℃湁澶寸華闃?br>
]]>
杈撳叆鍖呭惈澶氱粍嫻嬭瘯鏁版嵁銆傜涓涓暣鏁癗錛圢<=15錛?N琛ㄧず緇勬暟錛屾瘡緇勬暟鎹寘鍚袱涓暣鏁癮,b銆俛琛ㄧず涓涓崟浣嶇殑DNA涓茬殑琛屾暟錛宎涓哄鏁頒笖 3<=a<=39銆俠琛ㄧず閲嶅搴?1<=b<=20)銆?
Output:
杈撳嚭DNA鐨勫艦鐘訛紝姣忕粍杈撳嚭闂存湁涓絀鴻銆?
Sample Input:
2 3 1 5 4
Sample Output:
X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X
#include<iostream>
#include<string>
using namespace std;
int main()
{
int n; cin>>n;
int m=1;
for(int a,b;n&&cin>>a>>b;n--)
{ cout<<(m==1?"":"\n");
for(int i=1;i<=b;i++)
{for(int j=1;j<=a/2;j++)
{
cout<<string(j-1,' ');
cout<<'X';
cout<<string(a-2*(j-1)-2,' ');
cout<<'X'<<'\n';
}
cout<<string(a/2,' ');
cout<<'X'<<'\n';
for(int i=a/2;i>1;i--)
{ cout<<string(i-1,' ');
cout<<'X';
cout<<string(a-2*(i-1)-2,' ');
cout<<'X'<<'\n';
}
}
cout<<'X';cout<<string(a-2,' ');
cout<<'X'<<'\n';
m++;
}
}
]]>
{
int day;
for(day=1;day<=365;day++)
printf("I wish you happy everyday!\n");
}
]]>
#ifndef STRINT_HEADER
#define STRINT_HEADER
#include<iostream>
using namespace std;
//-------------------------------------
class StrInt{
enum { BYTENUM = 200 };
string _sign;
string _num;
public:
StrInt(const string& a = string("0") );
StrInt(const string& sign, const string& num);
friend StrInt mul( const StrInt& a, const StrInt& b );
friend StrInt add( const StrInt& a, const StrInt& b );
friend StrInt sub( const StrInt& a, const StrInt& b );
friend StrInt div( const StrInt& a, const StrInt& b );
friend StrInt mod( const StrInt& a, const StrInt& b );
friend istream& operator>>(istream& in, StrInt& a);
friend ostream& operator<<(ostream& out, const StrInt& a);
};//-----------------------------------
#endif // STRINT_HEADER
]]>
Time Limit:1000MS Memory Limit:10000K
Total Submit:2515 Accepted:1374
Description
Given a permutation of numbers from 1 to n, we can always get the sequence 1, 2, 3, ..., n by swapping pairs of numbers. For example, if the initial sequence is 2, 3, 5, 4, 1, we can sort them in the following way:
2 3 5 4 1
1 3 5 4 2
1 3 2 4 5
1 2 3 4 5
Here three swaps have been used. The problem is, given a specific permutation, how many swaps we needs to take at least.
Input
The first line contains a single integer t (1 <= t <= 20) that indicates the number of test cases. Then follow the t cases. Each case contains two lines. The first line contains the integer n (1 <= n <= 10000), and the second line gives the initial permutation.
Output
For each test case, the output will be only one integer, which is the least number of swaps needed to get the sequence 1, 2, 3, ..., n from the initial permutation.
Sample Input
2
3
1 2 3
5
2 3 5 4 1
Sample Output
0
3
]]>