题目链接:http://www.lydsy.com:808/JudgeOnline/problem.php?id=3085
题意:求n(<=10^100)之内最大的反素数。
思路:
优化2:
int prime[]=
{
1, 2, 3, 5, 7,
11, 13, 17, 19, 23,
29, 31, 37, 41, 43,
47, 53, 59, 61, 67,
71, 73, 79, 83, 89,
97, 101,103,107,109,
113,127,131,137,139,
149,151,157,163,167,
173,179,181,191,193,
197,199,211,223,227,
229,233,239,241,251
};
int K[]=
{
1,2,2,3,3,
4,4,5,5,5,
5,5,6,6,6,
6,6,6,6,7,
7,7,7,7,7,
7,7,7,7,7,
7,7,8,8,8,
8,8,8,8,8,
8,8,8,8,8,
8,8,8,8,8,
8,8,8,8,8
};
struct BIGINT
{
int a[27]; BIGINT(){}
BIGINT(char *s)
{
clr(a,0);
int i,L=strlen(s),cur=0;
for(i=L-1;i-3>=0;i-=4)
{
a[cur]=(s[i-3]-'0')*1000+
(s[i-2]-'0')*100+
(s[i-1]-'0')*10+
(s[i]-'0');
cur++;
}
if(i<0) return;
if(i==0) a[cur]=s[0]-'0';
else if(i==1) a[cur]=10*(s[0]-'0')+(s[1]-'0');
else if(i==2) a[cur]=100*(s[0]-'0')+10*(s[1]-'0')+(s[2]-'0');
}
BIGINT(int x)
{
clr(a,0);
a[0]=x;
} inline BIGINT operator*(int x)
{
int i;
BIGINT tmp;
for(i=0;i<27;i++) tmp.a[i]=a[i]*x;
for(i=0;i<26;i++)
{
tmp.a[i+1]+=tmp.a[i]/10000;
tmp.a[i]%=10000;
}
return tmp;
} int operator<(BIGINT p)
{
int i;
for(i=26;i>=0;i--)
{
if(a[i]<p.a[i]) return 1;
if(a[i]>p.a[i]) return 0;
}
return 0;
} int operator==(BIGINT p)
{
int i;
for(i=26;i>=0;i--)
{
if(a[i]!=p.a[i]) return 0;
}
return 1;
}
int operator<=(BIGINT p)
{
return *this==p||*this<p;
} void print()
{
int cur=26;
while(cur>0&&0==a[cur]) cur--;
printf("%d",a[cur]);
cur--;
while(cur>=0) printf("%04d",a[cur--]);
puts("");
}
}; char s[111];
BIGINT n;
int Max; int cnt2; BIGINT ans;
i64 ansFac; void DFS(int dep,BIGINT cur,i64 facNum,int preMax)
{
if(facNum>ansFac||facNum==ansFac&&cur<ans)
{
ans=cur;
ansFac=facNum;
}
int i;
i64 tmp=facNum;
int Min=min(preMax,2*K[Max]-1-1);
if(dep>1) Min=min(Min,cnt2/(K[dep]-1));
for(i=1;i<=Min;i++)
{
if(dep==1) cnt2=i;
cur=cur*prime[dep];
tmp+=facNum;
if(n<cur) break;
DFS(dep+1,cur,tmp,i);
}
} int main()
{ scanf("%s",s);
n=BIGINT(s); if(n==BIGINT(1))
{
puts("1");
return 0;
}
BIGINT cur=BIGINT(1);
while(cur<=n) cur=cur*prime[++Max];
DFS(1,BIGINT(1),1,100);
ans.print();
}