A：Repeating Characters
题意是给出一个重复次数量的R与一个字符串
输出重复后的字符串，很简单的一道题。
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
int t,c,s;
char ss[3000];
scanf("%d",&t);
while(t--)
{
scanf("%d %d %s",&c,&s,ss);
//cout<<c<<' '<<s<<' '<<ss<<endl;
int l = strlen(ss);
printf("%d ",c);
for(int i=0;i<l;i++)
{
for(int j=0;j<s;j++)
putchar(ss[i]);
}
printf("\n");
}
return 0 ;
}
BThe Rascal Triangle
题目给出一个三角形阵和对应数字的求解公式，求任意位置的数值是多少。
咋一看上去没啥规律，但是按照数组的格式写一下就会发现每一个竖行都是从1开始的等差数列，算一下等差比和项数就很容易就能求出来了。
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
#define Max 2505
#define FI first
#define SE second
#define ll long long
#define PI acos(-1.0)
#define inf 0x3fffffff
#define LL(x) ( x << 1 )
#define bug puts("here")
#define PII pair<int,int>
#define RR(x) ( x << 1 | 1 )
#define mp(a,b) make_pair(a,b)
#define mem(a,b) memset(a,b,sizeof(a))
#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )
using namespace std;
int main() {
int p,t;
int n,m;
scanf("%d",&p);
while(p--)
{
scanf("%d%d%d",&t,&n,&m);
if(n<0||m<0||n < m)
{
printf("%d 0\n",t);
continue;
}
long long ans = 1 + (n - m) * m;
printf("%d %lld\n",t,ans);
}
return 0 ;
}
CProgramming The EDSAC
C和D两道题正好凑成一对，一个是小数转二进制，另一个是二进制转小数。
因为最高涉及小数点后16位，直接用double很容易就失去精度，所以两道题都选择了模拟。
事实证明这种样子的模拟我还是能轻松处理的，就是因为这里给出的数都是完全按二进制存法来的（正数向下取整，负数向下取整）所以写得比较繁琐，一开始闹出不少问题。
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
#define Max 2505
#define FI first
#define SE second
#define ll long long
#define PI acos(-1.0)
#define inf 0x3fffffff
#define LL(x) ( x << 1 )
#define bug puts("here")
#define PII pair<int,int>
#define RR(x) ( x << 1 | 1 )
#define mp(a,b) make_pair(a,b)
#define mem(a,b) memset(a,b,sizeof(a))
#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )
using namespace std;
#define N 1111
int XX[17] = {0,0,0,0,0,1,5,2,5,8,7,8,9,0,6,2,5} ;
int YY[17] = {0,9,9,9,9,8,4,7,4,1,2,1,0,9,3,7,5} ;
string FK = "PQWERTYUIOJ#SZK*?F@D!HNM&LXGABCV" ;
int jian[17];
char s[20];
int ss[20];
int ans[20];
int l,q;
int main()
{
int t,c;
scanf("%d",&t);
while(t--)
{
scanf("%d %s",&c,&s);
memset(ss,0,sizeof(ss));
memset(ans,0,sizeof(ans));
l = strlen(s);
int i;
if(s[0] == '-')
{
i = 1;
ans[0] = 1;
}
else
{
i =0;
ans[0] = 0;
}
int j = 0;
ss[j] = s[i] - '0';
j++;
q = 0;
for(i = i + 2; i<l; i++)
{
ss[j] = s[i] - '0';
if(ss[j] > 0) q = 1;
j++;
}
if(q == 0)
{
if(ans[0] == 1) printf("%d ? 0 F\n",c);
else printf("%d P 0 F\n",c);
continue;
}
if(ss[0] > 0)
{
if((ss[0] ==1&&q == 1)|| ss[0] > 1)
{
printf("%d INVALID VALUE\n",c);
}
continue;
}
q = 0;
if(s[0] == '-')
{
memset(jian,0,sizeof(jian));
jian[0] = 1;
for(i=16; i>=0; i--)
{
ss[i] = jian[i] - ss[i];
if(ss[i] < 0)
{
ss[i] = ss[i] + 10;
jian[i-1] = jian[i-1] - 1;
}
}
q = 1;
}
for(i = 1; i <= 16; i++)
{
for(j = 16; j > 0; j--)
{
ss[j] =ss[j] * 2;
}
for(j = 16; j > 0; j--)
{
if(ss[j] >= 10)
{
ss[j] = ss[j] - 10;
ss[j-1]++;
}
}
if(ss[0] == 1)
{
ss[0] = 0;
ans[i] = 1;
if(ans[i] == 2) ans[i] = 1;
}
else
{
ans[i] = 0;
}
}
if(q == 1)
{
q = 0;
for(i=0; i<17; i++)
{
if(ss[i] > 0)
{
q = 1;
break;
}
}
if(q == 1)
{
ans[16] += 1;
int i = 16;
while(i >= 0 && ans[i] > 1)
{
ans[i] = ans[i] - 2;
i--;
ans[i]++;
}
}
}
int cc = 0,tt = 0;
char fr;
for(i=0; i<5; i++)
{
cc = cc * 2 + ans[i];
}
for(i=5; i<16; i++)
{
tt = tt * 2 + ans[i];
}
if(ans[16] == 0) fr = 'F';
else fr = 'D';
printf("%d %c %d %c\n",c,FK[cc],tt,fr);
}
return 0 ;
}
DDecoding EDSAC Data
写C的时候我只把加减法模拟出来了，写到D的时候干脆把二进制一位小数到16为小数都模拟出来直接加。
只能说是习惯去模拟了，还是简单暴力的好~~^_^~~
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int ans[17];
int cal[17];
int jian[17];
int x[16][17] = {0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,2,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,2,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,2,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,3,1,2,5,0,0,0,0,0,0,0,0,0,0,0,0,0,1,5,6,2,5,0,0,0,0,0,0,0,0,0,0,0,0,0,7,8,1,2,5,0,0,0,0,0,0,0,0,0,0,0,0,3,9,0,6,2,5,0,0,0,0,0,0,0,0,0,0,0,1,9,5,3,1,2,5,0,0,0,0,0,0,0,0,0,0,0,9,7,6,5,6,2,5,0,0,0,0,0,0,0,0,0,0,4,8,8,2,8,1,2,5,0,0,0,0,0,0,0,0,0,2,4,4,1,4,0,6,2,5,0,0,0,0,0,0,0,0,1,2,2,0,7,0,3,1,2,5,0,0,0,0,0,0,0,0,6,1,0,3,5,1,5,6,2,5,0,0,0,0,0,0,0,3,0,5,1,7,5,7,8,1,2,5,0,0,0,0,0,0,1,5,2,5,8,7,8,9,0,6,2,5} ;
string FK = "PQWERTYUIOJ#SZK*?F@D!HNM&LXGABCV" ;
char sh,fr;
int zh,q;
int main()
{
/*for(int i=0;i<16;i++)
{
for(int j=0;j<17;j++)
{
cout<<x[i][j];
}
cout<<endl;
}*/
int t,c,tt;
scanf("%d",&t);
while(t--)
{
scanf("%d %c %d %c",&c,&sh,&zh,&fr);
//cout<<c<<' '<<sh<<' '<<zh<<' '<<fr<<endl;
memset(ans,0,sizeof(ans));
memset(cal,0,sizeof(cal));
for(int i=0; i<32; i++)
{
if(FK[i] == sh)
{
tt = i;
}
}
if(fr == 'F') cal[16] = 0;
else cal[16] = 1;
int i = 4;
while(tt != 0)
{
cal[i] = tt % 2;
tt = tt / 2;
i--;
}
i = 15;
tt = zh;
while(tt != 0)
{
cal[i] = tt % 2;
tt = tt / 2;
i--;
}
/*for(int i=0;i<17;i++)
{
cout<<cal[i];
}
cout<<endl;*/
for(i=1; i<17; i++)
{
if(cal[i] == 1)
{
for(int j=16; j>=0; j--)
{
ans[j] =ans[j] + x[i-1][j];
if(ans[j] >= 10)
{
ans[j] = ans[j] - 10;
ans[j-1] += 1;
}
}
//cout<<'a';
}
}
//cout<<endl;
if(cal[0] == 1)
{
memset(jian,0,sizeof(jian));
jian[0] = 1;
for(i=16; i>=0; i--)
{
ans[i] = jian[i] - ans[i];
if(ans[i] < 0)
{
ans[i] = ans[i] + 10;
jian[i-1] = jian[i-1] - 1;
}
}
}
/*for(i=0; i<17; i++)
{
cout<<ans[i];
}
cout<<endl;*/
printf("%d ",c);
if(cal[0] == 1) printf("-");
printf("%d.%d",ans[0],ans[1]);
i = 16;
while(i != 0 && ans[i] == 0) i--;
int j = 2;
while(j <= i)
{
printf("%d",ans[j]);
j++;
}
printf("\n");
}
return 0 ;
}
HMaximum in the Cycle of 1
求按照题目给出的计算方法所得出的循环中所得值为k的种数有多少因为n<=20直接各种搜时间消耗为20！肯定超时，所以推规律，得出n和k之间的关系，然后直接按规律计算就行了。
虽说我按位按样例推出了规律，但是很多细节上的东西还是稀里糊涂的。回头再推一下规律是怎么出来的。（按照需求数排列方式数*可以插空的数量*剩余空当的排列数来计算，虽说推出来需求排列数是k！，但是想不通为何，很奇怪。。）
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
#define Max 2505
#define FI first
#define SE second
#define ll long long
#define PI acos(-1.0)
#define inf 0x3fffffff
#define LL(x) ( x << 1 )
#define bug puts("here")
#define PII pair<int,int>
#define RR(x) ( x << 1 | 1 )
#define mp(a,b) make_pair(a,b)
#define mem(a,b) memset(a,b,sizeof(a))
#define REP(i,s,t) for( int i = ( s ) ; i <= ( t ) ; ++ i )
using namespace std;
ll s[22];
int c,n,k;
void get()
{
s[0] = s[1] = 1;
for(int i=2;i<=20;i++)
{
s[i] = s[i-1] * i;
}
}
long long cal()
{
if(k == 1) return s[n-1];
ll ans = 0;
ll sum;
int kk = k - 2;
for(int i = k;i > 1;i--)
{
sum = 1;
for(int j = 0;j < i - 2;j++)
{
sum = sum * (kk - j) / (j+1);
}
//cout<<sum<<endl;
ans += sum * s[n-i] * s[i-1];
}
return ans;
}
int main() {
get();
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&c,&n,&k);
printf("%d %lld\n",c,cal());
}
return 0 ;
}