/* 合唱队
#include <stdio.h>
int main()
{
int n;
int i,j;
int a[100];
int f1[100];//存放合唱队的人数（从左向右）
int f2[100];//存放合唱队的人数（从右向左）
int ans = 0;
//cin>>n;
scanf("%d",&n);
for(i=1;i<=n;i++)//第0个位置不存放数据，符合平常的思维习惯。
//cin>>a[i];
scanf("%d",&a[i]);
for(i=1;i<=n;i++)//由左向右依次遍历
{
f1[i] = 1;//至少有一个人符合条件，就是他自己。所以赋初值1.
for(j=1;j<i;j++)
{
if(a[i]>a[j]&&f1[i]<f1[j]+1) //f1[i]<f1[j]+1很关键的条件，动态问题
f1[i]= f1[j]+1;
}
}
for(i=n;i>=1;i--)//由右向左依次遍历
{
f2[i] = 1;//至少有一个人符合条件，就是他自己。所以赋初值1.
for (j=i+1;j<=n;j++)
{
if(a[i]>a[j]&&f2[i]<f2[j]+1)
f2[i]=f2[j]+1;
}
}
for(i=1;i<=n;i++)
if(ans<f1[i]+f2[i]-1)
ans=f1[i]+f2[i]-1;//最佳解
//cout<<n-ans<<endl;//需要出列的人数
printf("%d",n-ans);
return 0;
}
*/
/* 坐标移动
#include <iostream>
using namespace std;
void main()
{
int x=0,y=0;
int i,j,k;
char a[100],b[30][5];
j=0;
k=0;
gets(a);
for (i=0;i<strlen(a);i++)
{
if (a[i] != ';')
{
b[j][k] = a[i];
k++;
}
else
{
b[j][k]=' ';
j++;
k=0;
}
}
for (i=0;i<j;i++)
{
k=strlen(b[i]);
if (k==2)
{
if (b[i][0]=='A'&& (b[i][1]>='0'&&b[i][1]<='9'))
{
x = x-(b[i][1]-'0');
}
if (b[i][0]=='D'&& (b[i][1]>='0'&&b[i][1]<='9'))
{
x = x+(b[i][1]-'0');
}
if (b[i][0]=='W'&& (b[i][1]>='0'&&b[i][1]<='9'))
{
y = y+(b[i][1]-'0');
}
if (b[i][0]=='S'&& (b[i][1]>='0'&&b[i][1]<='9'))
{
y = y-(b[i][1]-'0');
}
}
else if(k==3)
{
if (b[i][0]=='A'&& (b[i][1]>='0'&&b[i][1]<='9')&&(b[i][2]>='0'&&b[i][2]<='9'))
{
x = x-((b[i][1]-'0')*10+b[i][2]-'0');
}
if (b[i][0]=='D'&& (b[i][1]>='0'&&b[i][1]<='9')&&(b[i][2]>='0'&&b[i][2]<='9'))
{
x = x+((b[i][1]-'0')*10+b[i][2]-'0');
}
if (b[i][0]=='W'&& (b[i][1]>='0'&&b[i][1]<='9')&&(b[i][2]>='0'&&b[i][2]<='9'))
{
y = y+((b[i][1]-'0')*10+b[i][2]-'0');
}
if (b[i][0]=='S'&& (b[i][1]>='0'&&b[i][1]<='9')&&(b[i][2]>='0'&&b[i][2]<='9'))
{
y = y-((b[i][1]-'0')*10+b[i][2]-'0');
}
}
}
cout<<x<<","<<y<<endl;
}
*/
/* 报数
#include <iostream>
using namespace std;
struct Node
{
int Data;
Node *next;
};
class list
{
public:
Node *head;
list() {head = NULL;}
void InsertNode(int a);
};
void list::InsertNode(int a)
{
Node *p,*s,*q;
s = (Node *)new(Node);
s->Data = a;
p = head;
q = head;
if (p == NULL)
{
head = s;
s->next = s;
}
else
{
while (q->next!=p)
{
q = q->next;
}
q->next = s;
s->next = p;
}
}
void main()
{
int n,i;
char s;
list slist;
Node *p,*q;
do
{
scanf("%d",&n);
p = NULL;
q = NULL;
slist = list();
for (i=1;i<=n;i++)
{
slist.InsertNode(i);
}
p = slist.head;
while (p!=q)
{
for (i=1;i<3;i++)
{
q = p;
p = p->next;
}
q->next = p->next;
delete p;
p = q->next;
}
cout<<p->Data<<endl;
}
while(scanf("%c",&s) != EOF);
}
*/
/* 数字基ROOT*/
/*
#include <iostream>
using namespace std;
int fun1(char *a)
{
int i;
int sum = 0;
for (i=0;i<strlen(a);i++)
{
if (a[i]>='0'&&a[i]<='9')
{
sum = sum + a[i] -'0';
}
else
return -1;
}
return sum;
}
int fun2(int a)
{
if (a<10) return a;
int temp;
int sum = 0;
while (a)
{
temp = a%10;
a=a/10;
sum = sum + temp;
}
return sum;
}
void main()
{
char a[100];
int sum;
while (gets(a))
{
sum = fun1(a);
while(sum>9)
{
sum = fun2(sum);
}
cout<<sum<<endl;
}
}
*/
/*识别有效IP地址和掩码
#include <iostream>
#include <string>
#include <vector>
using namespace std;
static int result[7]={0}; //保存最后结果的数组
string Chang(string str) //转换为二进制
{
int num=atoi(str.c_str());
string tmp;
char tmp1[10]={0};
itoa(num,tmp1,2);
tmp=string(tmp1);
if (tmp.size()<8)
{
string tmp2(8-tmp.size(),'0');
str=tmp2+tmp;
}
else
str=tmp;
return str;
}
void checkNetSegment(string mask,string ip1)
{
int count1=0,count2=0;
int count4=0,count5=0;
string str1[4],str3[4];
for(int i=0; i<mask.size(); i++) //掩码的点个数(不连续的点等于3个)
if(mask[i]=='.' && mask[i+1]!='.')
count1++;
for(int i=0; i<ip1.size(); i++) //ip1的点个数
if(ip1[i]=='.' && ip1[i+1]!='.')
count2++;
for(int i=0; i<mask.size(); i++) //掩码的点个数(所有的点等于3个)
if(mask[i]=='.')
count4++;
for(int i=0; i<ip1.size(); i++) //ip1的点个数
if(ip1[i]=='.')
count5++;
if(count1==count2&&count1==3
&& count4==count5&&count4==3) //如果格式正确
{
int i=0; //ip1先分解为4段
int flag1=0;
int k=0;
while(i<ip1.size())
{
int tmp=i;
while(isdigit(ip1[tmp]))
tmp++;
str1[k]=ip1.substr(i,tmp-i);
k++;
i=tmp+1;
}
i=0; //掩码分为4段
int flag3=0,flagmk=0;
k=0;
string test;
while(i<mask.size())
{
int tmp=i;
while(isdigit(mask[tmp]))
tmp++;
str3[k]=mask.substr(i,tmp-i);
k++;
i=tmp+1;
}
for(int i=0; i<4; i++) //判断每一段的值是否在0-255之间
if(atoi(str1[i].c_str())<256 &&atoi(str1[i].c_str())>=0)
flag1++;
for(int i=0; i<4; i++)
if(atoi(str3[i].c_str())<256 &&atoi(str3[i].c_str())>=0)
flag3++;
for (int i=0;i<4;i++) //判掩码是不是1在0前面
{
string temp=Chang(str3[i]);
test+=temp;
}
for (int j=1;j<32-1;j++)
if (test[j-1]>test[j] && test[j]<test[j+1])
{
flagmk=1;
break;
}
if(flag1==flag3 &&flag3==4 &&flagmk==0) //格式正确
{
if (str1[0]>="1" &&str1[0]<="126" )
result[0]++;
else if (str1[0]>="128" &&str1[0]<="191")
result[1]++;
else if (str1[0]>="192"&&str1[0]<="223")
result[2]++;
else if (str1[0]>="224"&&str1[0]<="239")
result[3]++;
else if (str1[0]>="240"&&str1[0]<="255")
result[4]++;
else if (str1[0]=="127")
result[5]++;
if (str1[0]=="10"|| //私有IP
(str1[0]=="172"&&str1[1]>="16"&&str1[1]<="31")||
(str1[0]=="192"&&str1[1]=="168"))
result[6]++;
}
else
result[5]++;
}
else //格式不正确
result[5]++;
}
int main()
{
string s1,ms,ip;
while (getline(cin,s1))
{
int position=s1.find("~",0);
ip=s1.substr(0,position);
ms=s1.substr(position+1,s1.size()-1);
checkNetSegment(ms,ip);
}
for (int i=0;i<7;i++)
{
cout<<result[i];
if (i==6)
cout<<endl;
else
cout<<" ";
}
return 0;
}
*/
/*mimajiaoyan*/
#include <iostream>
using namespace std;
int LRS(char *a)
{
int leng=strlen(a);
int Max=0,count=0;
int start1,start2;
for(int i=0;i<leng;i++)
for(int j=i+1;j<leng;j++)
{
start1=i;start2=j;
while(a[start1]==a[start2] && start1<leng && start2<leng)
{
count++;
start1++;
start2++;
}
if(count>Max)
Max=count;
count=0;
}
return Max;
}
int check(char *a)
{
int flag[4]={0};
int i,sum=0;
if (strlen(a) < 8)
{
return 0;
}
for(i=0;i<strlen(a);i++)
{
if (a[i]>='A'&&a[i]<='Z')
{
flag[0] = 1;
}
else if(a[i]>='a'&&a[i]<='z')
{
flag[1] = 1;
}
else if(a[i]>='0'&&a[i]<='9')
{
flag[2] = 1;
}
else
flag[3] = 1;
}
for(i=0;i<4;i++)
{
sum = sum + flag[i];
}
if (sum>=3)
{
if (LRS(a)>2)
return 0;
else
return 1;
}
else
return 0;
}
void main()
{
char a[20];
while (gets(a))
{
if (check(a) == 0)
cout<<"NG"<<endl;
else
cout<<"OK"<<endl;
}
}
中级题和部分初级题