P1603 斯诺登的密码 - 华师一附中OI组

2013年X月X日，俄罗斯办理了斯诺登的护照，于是他混迹于一架开往委内瑞拉的飞机。但是，这件事情太不周密了，因为FBI的间谍早已获悉他的具体位置——但这不是最重要的——最重要的是如果要去委内瑞拉，那么就要经过古巴，而经过古巴的路在美国的掌控之中。丧心病狂的奥巴马迫降斯诺登的飞机，搜查时却发现，斯诺登杳无踪迹。但是，在据说是斯诺登的座位上，发现了一张纸条。纸条由纯英文构成：Obama is a two five zero.（以"."结束输出，只有6个单词+一个句号，句子开头如没有大写亦为合法）这句话虽然有点无厘头，但是警官陈珺骛发现这是一条极其重要的线索。他在斯诺登截获的一台笔记本中找到了一个C++程序，输入这条句子后立马给出了相对应的密码。陈珺鹜高兴得晕了过去，身为警官的你把字条和程序带上了飞机，准备飞往曼哈顿国际机场，但是在飞机上检查的时候发现——程序被粉碎了！飞机抵达华盛顿只剩5分钟，你必须在这5分钟内编写（杜撰）一个程序，免受上司的10000000000%10大板。破译密码的步骤如下：

#include<iostream>
using namespace std;
string s,s1[26]={" one "," two "," three "," four "," five "," six "," seven "," eight "," nine "," ten "," eleven "," twelve "," thirteen ",
" fourteen "," fifteen "," sixteen "," seventeen "," eighteen "," nineteen "," twenty "," a "," both "," another "," first "," second "," third "};
int a[20],num[10];
long long int ans=99999999999,t=0;
bool visit[10];
void dfs(int x,int y)
{
if(x>y)
{
ans=min(ans,t);
return;
}
for(int i=0;i<=y;i++)
{
if(!visit[i])
{
visit[i]=1;
t=t*100+num[i];
dfs(x+1,y);
t=(t-num[i])/100;
visit[i]=0;
}
}
}
int main()
{
getline(cin,s);
s[s.size()-1]=' ';
s=" "+s;
/// cout<<s<<endl;;
for(int i=0;i<=25;i++)
{
/// cout<<s1[i];
int j=0;
j=s.find(s1[i],0);
/// cout<<j<<" ";
if(i<=19)a[i]++;
if(i==20 || i==22 || i==23)a[0]++;
if(i==21 || i==24)a[1]++;
if(i==25)a[2]++;
while(j!=-1)
{
j=s.find(s1[i],j+5);
/// cout<<j<<" ";
if(i<=19)a[i]++;
if(i==20 || i==22 || i==23)a[0]++;
if(i==21 || i==24)a[1]++;
if(i==25)a[2]++;
}
/// cout<<endl;
if(i<=19)a[i]--;
if(i==20 || i==22 || i==23)a[0]--;
if(i==21 || i==24)a[1]--;
if(i==25)a[2]--;
}
/* for(int i=0;i<=19;i++)
cout<<a[i]<<" ";
cout<<endl;*/
int n=0;
for(int i=0;i<=19;i++)
for(int j=1;j<=a[i];j++)
num[n++]=((i+1)*(i+1))%100;
n--;
dfs(0,n);
cout<<ans<<endl;
return 0;
}

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#include<iostream>
#include<map>
using namespace std;
map<string,int>m;
string s,ans[6];
int l,t;
int main()
{
m["one"]=1;m["two"]=2;m["three"]=3;m["four"]=4;m["five"]=5;
m["six"]=6;m["seven"]=7;m["eight"]=8;m["nine"]=9;m["ten"]=10;
m["eleven"]=11;m["twelve"]=12;m["thirteen"]=13;m["fourteen"]=14;m["fifteen"]=15;
m["sixteen"]=16;m["seventeen"]=17;m["eighteen"]=18;m["nineteen"]=19;m["twenty"]=20;
m["a"]=1;m["both"]=2;m["another"]=1;m["first"]=1;m["second"]=2;m["third"]=3;
for(int i=0;i<6;i++)
{
cin>>s;
l=s.size();
for(int j=0;j<l;j++)if('A'<=s[j]&&s[j]<='Z')s[j]+='a'-'A';
if(m[s])
{
int k=m[s]*m[s]%100;
ans[t]+=k/10+'0';
ans[t]+=k%10+'0';
t++;
}
}
if(t==0){cout<<'0'<<endl;return 0;}
sort(ans,ans+t);
if(ans[0][0]=='0')cout<<ans[0][1];
else cout<<ans[0];
for(int i=1;i<6;i++)cout<<ans[i];cout<<endl;
return 0;
}

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