A. Display The Number
answer
The bigger the better number of questions asked, and the answer will have hinted burst longlong,
so we have the following policies:
\ (the n-\) is even full-fill \ (1 \) , because \ (1 \) only two cells, it is worth
\ (n \) is odd, then the first mad fill \ (1 \) , and then left \ (3 \) a time to fill a \ (7 \) just fine
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
typedef long long LL;
typedef pair<int,int> PII;
#define X first
#define Y second
inline int read()
{
int x=0,f=1;char c=getchar();
while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
while(isdigit(c)){x=x*10+c-'0';c=getchar();}
return x*f;
}
int w[10]={6,2,5,5,4,5,6,3,7,6};
int T,n;
int main()
{
T=read();
while(T--)
{
n=read();
if(n%2==0)
for(int i=1;i<=n/2;i++)printf("1");
else
{
printf("7");
for(int i=1;i<n/2;i++)printf("1");
}
printf("\n");
}
return 0;
}B. Infinite Prefixes
answer
Essay card for a long time, think how hard to describe language solution
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
typedef long long LL;
typedef pair<int,int> PII;
#define X first
#define Y second
inline int read()
{
int x=0,f=1;char c=getchar();
while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
while(isdigit(c)){x=x*10+c-'0';c=getchar();}
return x*f;
}
const int maxn=100010;
int T,n,x,p0[maxn],p1[maxn],pre[maxn],MAX;
char s[maxn];
int main()
{
T=read();
while(T--)
{
mem(p0,0);mem(p1,0);mem(pre,0);
n=read();x=read();
int cnt=0;
scanf("%s",s+1);
for(int i=1;i<=n;i++)p0[i]=p0[i-1]+(s[i]=='0'),p1[i]=p1[i-1]+(s[i]=='1'),pre[i]=p0[i]-p1[i];
if(x==0)cnt=1;
if(pre[n])
{
for(int i=1;i<=n;i++)if((x-pre[i])%pre[n]==0 && (x-pre[i])/pre[n]>=0)cnt++;
}
else
{
for(int i=0;i<=n;i++)if(x==pre[i])cnt++;
}
if(!pre[n] && cnt)printf("-1\n");
else printf("%d\n",cnt);
}
return 0;
}C. Obtain The String
answer
Especially as the string match
but I did not do it, the template string pretreatment, save string matching delays.
Set \ (pre [i] \) represents the letter \ (I \) is the earliest position, \ (SUF [I] [J] \) represents shown in section \ (I \) after characters, letters \ (J \ ) earliest position, this can be \ (O (26n) \) pretreated out
and then jump directly to repeatedly cross at the top around it
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
typedef long long LL;
typedef pair<int,int> PII;
#define X first
#define Y second
inline int read()
{
int x=0,f=1;char c=getchar();
while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
while(isdigit(c)){x=x*10+c-'0';c=getchar();}
return x*f;
}
const int maxl=100010;
int T;
char s[maxl],t[maxl];
int pre[30],suf[maxl][30];//pre(i)表示字符i最早出现位置 ，suf(i,j)表示在i字符后，字符j最早出现位置
int main()
{
T=read();
while(T--)
{
mem(pre,42);mem(suf,42);
scanf("%s%s",s,t);
int l1=strlen(s),l2=strlen(t),cnt=0,ok=0;
for(int i=0;i<l1;i++)pre[s[i]-'a']=min(pre[s[i]-'a'],i);
for(int i=0;i<l1-1;i++)suf[i][s[i+1]-'a']=i+1;
for(int i=l1-1;i>=0;i--)
for(int j=0;j<26;j++)suf[i][j]=min(suf[i][j],suf[i+1][j]);
for(int i=0;i<l2;)
{
++cnt;
int pos=pre[t[i]-'a'];i++;
if(pos>=l1){ok=1;break;}
while(suf[pos][t[i]-'a']<l1)pos=suf[pos][t[i]-'a'],i++;
}
if(!ok)printf("%d\n",cnt);
else printf("%d\n",-1);
}
return 0;
}D. Same GCDs
answer
Set \ (GCD (A, m) = X \) , there are \ (a = k_1x, m =
k_2x \) may know \ (GCD (A + B, m) = X \) , there are \ (a + b = k_3x \)
the question now becomes, in the closed interval \ ([k_1, k_1 + k_2-1 ] \) the number of inner (K_2 \) \ prime
split into prefix problem, \ (n-\) after the decomposition of the quality factor to the principle of inclusion and exclusion factors
have a similar problem this is the link
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<queue>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
typedef long long LL;
typedef pair<int,int> PII;
#define X first
#define Y second
inline LL read()
{
LL x=0,f=1;char c=getchar();
while(!isdigit(c)){if(c=='-')f=-1;c=getchar();}
while(isdigit(c)){x=x*10+c-'0';c=getchar();}
return x*f;
}
int T;
LL a,m,x,k1,k2;
LL gcd(LL a,LL b){return b==0 ? a : gcd(b,a%b);}
#include<vector>
vector<LL> pme;
LL count_prime(LL x,LL n){
pme.clear();
LL i,j;
for(i=2;i*i<=n;i++)
if(n%i==0){
pme.push_back(i);
while(n%i==0)n/=i;
}
if(n>1)pme.push_back(n);
LL sum=0,value,cnt;
for(i=1;i<(1<<pme.size());i++){
value=1;
cnt=0;
for(j=0;j<pme.size();j++){
if(i&(1<<j)){
value*=pme[j];
cnt++;
}
}
if(cnt&1)
sum+=x/value;
else sum-=x/value;
}
return x-sum;
}
int main()
{
T=(int)read();
while(T--)
{
a=read();m=read();
x=gcd(a,m);
k1=a/x;k2=m/x;
cout<<count_prime(k2+k1-1,k2)-count_prime(k1-1,k2)<<endl;
}
return 0;
}nonsense
The brain or not, deal with the problem is too slow ah