Section 1.1
| PROB Broken Necklace |
这里也是巧解……
/*
ID:liwei101
PROG:beads
LANG:C++
*/
#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <list>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <set>
#define PI acos(-1.0)
#define mem(a,b) memset(a,b,sizeof(a))
#define sca(a) scanf("%d",&a)
#define pri(a) printf("%d\n",a)
#define MM 1000002
#define MN 2005
#define INF 168430090
using namespace std;
typedef long long ll;
int main()
{
ifstream fin("beads.in");
ofstream fout("beads.out");
int n,i,go=0,pre=0,w=0,sum=0;
string s;
char c=‘ ‘;
fin>>n>>s;
s+=s; //圆圈字符串可以把其变为两个字母串连接，首尾相连即可……
for(i=0;i<n*2;i++)
{
if(s[i]==‘w‘) go++,w++;
else if(s[i]==c) go++,w=0;
else
{
sum=max(go+pre,sum); //两端项链之和
pre=go-w; //左端就是右端减去有字母w的个数
go=w+1; //右端往前刚开始的个数就是前面有字母w个数加上当前的一个
w=0; //w个数重置为0
c=s[i]; //当前字母
}
}
sum=max(go+pre,sum);
fout<<(sum>n?n:sum)<<endl;
return 0;
}
|
Section 1.2 PROB Milking Cows |
/*
ID:liwei101
PROG:milk2
LANG:C++
*/
#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <list>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <set>
#define PI acos(-1.0)
#define mem(a,b) memset(a,b,sizeof(a))
#define sca(a) scanf("%d",&a)
#define pri(a) printf("%d\n",a)
#define MM 1000002
#define MN 2005
#define INF 168430090
using namespace std;
typedef long long ll;
int hash[MM],s[3];
int main()
{
ifstream fin("milk2.in");
ofstream fout("milk2.out");
int n,i,j,a,b,Min=MM,Max=0;
fin>>n;
for(i=0; i<n; i++)
{
fin>>a>>b;
b--;
Min=min(Min,a);
Max=max(Max,b);
for(j=a; j<=b; j++)
hash[j]=1; //把出来过的标记为1
}
for(i=Min; i<Max; i+=j)
{
for(j=0; hash[i+j]==hash[i]; j++); //这句话就是为什么刚开始得：b--了，如果不减的话，这两句循环都得再减1才行……
s[hash[i]]=max(s[hash[i]],j); //哈希取址的思想真的太强大了……
}
fout<<s[1]<<‘ ‘<<s[0]<<endl;
return 0;
}
/*
ID:liwei101
PROG:namenum
LANG:C++
*/
#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <list>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <set>
#define PI acos(-1.0)
#define mem(a,b) memset(a,b,sizeof(a))
#define sca(a) scanf("%d",&a)
#define pri(a) printf("%d\n",a)
#define MM 1000002
#define MN 2005
#define INF 168430090
using namespace std;
typedef long long ll;
int main()
{
ifstream fin("namenum.in");
ofstream fout("namenum.out");
ifstream dict("dict.txt");
string s,ss;
int a[27],flag=1;
a[‘A‘-‘A‘]=a[‘B‘-‘A‘]=a[‘C‘-‘A‘]=2;
a[‘D‘-‘A‘]=a[‘E‘-‘A‘]=a[‘F‘-‘A‘]=3;
a[‘G‘-‘A‘]=a[‘H‘-‘A‘]=a[‘I‘-‘A‘]=4;
a[‘J‘-‘A‘]=a[‘K‘-‘A‘]=a[‘L‘-‘A‘]=5;
a[‘M‘-‘A‘]=a[‘N‘-‘A‘]=a[‘O‘-‘A‘]=6;
a[‘P‘-‘A‘]=a[‘R‘-‘A‘]=a[‘S‘-‘A‘]=7;
a[‘T‘-‘A‘]=a[‘U‘-‘A‘]=a[‘V‘-‘A‘]=8;
a[‘W‘-‘A‘]=a[‘X‘-‘A‘]=a[‘Y‘-‘A‘]=9;
fin>>ss;
while(dict>>s)
{
string sss;
for(int i=0;i<s.size();i++)
{
sss+=a[s[i]-‘A‘]+‘0‘;
}
if(sss==ss) {fout<<s<<endl;flag=0;}
}
if(flag) fout<<"NONE"<<endl;
return 0;
}
/*
PROG:palsquare
LANG:C++
ID:liwei101
*/
#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <list>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define PI acos(-1.0)
#define mem(a,b) memset(a,b,sizeof(a))
#define sca(a) scanf("%d",&a)
#define pri(a) printf("%d\n",a)
#define f(i,a,n) for(i=a;i<n;i++)
#define F(i,a,n) for(i=a;i<=n;i++)
#define MM 1000002
#define MN 5005
#define INF 1000000007
using namespace std;
typedef long long ll;
char fuck[22]={‘0‘, ‘1‘, ‘2‘, ‘3‘, ‘4‘, ‘5‘, ‘6‘, ‘7‘, ‘8‘, ‘9‘, ‘A‘, ‘B‘, ‘C‘, ‘D‘, ‘E‘, ‘F‘, ‘G‘, ‘H‘, ‘I‘, ‘J‘};
string convert(int i,int b)
{
string a;
while(i)
{
a+=fuck[i%b];
i/=b;
}
for(int i=0,j=a.size()-1;i<j;i++,j--)
{
char c=a[i];
a[i]=a[j];
a[j]=c;
}
return a;
}
int cmp(string s)
{
int i,j;
for(i=0,j=s.size()-1;i<j;i++,j--)
if(s[i]!=s[j]) return 0;
return 1;
}
int main()
{
ofstream fout("palsquare.out");
ifstream fin("palsquare.in");
int i,b;
fin>>b;
for(i=1;i<=300;i++)
{
string s1=convert(i,b);
string s2=convert(i*i,b);
if(cmp(s2)) fout<<s1<<‘ ‘<<s2<<endl;
}
return 0;
}
服务器编程入门（7）I/O复用,布布扣,bubuko.com
原文：http://blog.csdn.net/zs634134578/article/details/19929449