B LaTeX Expert
It’s not difficult to think, but it’s a bit disgusting to read, use gets getsg e t s read one line, then usestrcmp strcmps t r c m p to determine whether it is a blank line
#include <bits/stdc++.h>
#define ll long long
#define pi pair<int,int>
#define mk make_pair
#define pb push_back
using namespace std;
const int maxn = 1e5+100;
char a[maxn],A[maxn];
struct node{
string s;
int id;
}t[maxn];
struct node1{
char s[maxn];
int id;
bool operator < (const node1 & t)const
{
return id < t.id;
}
}tmp[105];
map<string,int>mp;
int main()
{
int cnt = 0,flag = 0;
while(gets(a)){
if(strcmp("\\begin{thebibliography}{99}", a) == 0)
{
break;
}
// printf("*%s*",a);
for(int i=0;i<strlen(a);i++)
{
if(a[i] == '{')
{
if(flag == 0)flag = 1;
}
else if(a[i] == '}')flag = 0,++cnt;
else if(flag == 1)
{
t[cnt].s += a[i];
}
}
}
for(int i=0;i<cnt;i++)
{
mp[t[i].s] = i;
}
int tot = 0;
flag = 0;
string ttt="";
// puts("DDD!!!!!!!!!!!!!!!!!!!!!");
while(gets(a))
{
if(strcmp("\\end{thebibliography}", a) == 0)
{
break;
}
if(strcmp(a , "")==0)continue;
strcpy(tmp[++tot].s,a);
// printf("*");
// puts(tmp[tot].s);
for(int i=0;i<strlen(tmp[tot].s);i++)
{
if(tmp[tot].s[i] == '{')
{
ttt = "";
if(flag == 0)flag = 1;
}
else if(tmp[tot].s[i] == '}')
{
flag = 0;
tmp[tot].id = mp[ttt];
}
else if(flag == 1)
{
ttt += tmp[tot].s[i];
}
}
}
// for(int i=1;i<=tot;i++)cout<<tmp[i].s<<' '<<tmp[i].id<<'\n';
cnt = 0;
flag = 1;
for(int i=1;i<=tot;i++)
{
if(tmp[i].id != cnt++)
{
flag = 0;
break;
}
}
if(flag == 0)
{
puts("Incorrect");
sort(tmp+1,tmp+tot+1);
puts("\\begin{thebibliography}{99}");
for(int i=1;i<=tot;i++)printf("%s\n",tmp[i].s);
puts("\\end{thebibliography}");
}
else puts("Correct");
return 0;
}
I Minimal Product
Question meaning: first find AA by adding to the questionA array, and then find thati <j i <ji<j && a [ i ] < a [ j ] a[i]<a[j] a[i]<a[j] && a [ i ] ∗ a [ j ] a[i]*a[j] a[i]∗a The minimum value of [ j ] .
Problem solution: askAAIn the case of A array, there is a place to take the modulus of 2 to the 32nd power. We can useunsigned − int unsigned -intunsigned−i n t takes the modulus automatically, because its range is 0-2^32-1, and the automatic overflow when calculating is equivalent to taking the modulus. That condition maintains the minimum prefix for positive numbers. For negative numbers, maintaining the maximum value of the prefix smaller than it means maintaining the maximum value of the suffix backwards.
#include <bits/stdc++.h>
#define ll long long
#define pi pair<int,int>
#define mk make_pair
#define pb push_back
using namespace std;
const int maxn = 1e7+10;
unsigned int b[maxn];
ll a[maxn];
ll ans[maxn];
int main()
{
int T;
cin >> T;
while(T--)
{
ll n,l,r;
unsigned int x,y,z;
cin >> n >> l >> r >> x >> y >> z >> b[1] >> b[2];
for(int i=3;i<=n;i++)
{
b[i] = (x*b[i-2] + b[i-1]*y + z);
}
for(int i=1;i<=n;i++)
{
a[i] = b[i]%(r-l+1)+l;
// printf("%d ",a[i]);
}
ll mi = 1e18,ma = -1e18;
ll ans = 4e18;
for(int i=1;i<=n;i++)
{
if(a[i] >= 0 && a[i] > mi)ans = min(ans , a[i]*mi);
mi = min(a[i],mi);
}
for(int i = n;i >=1;i--)
{
if(a[i] <= 0)
{
if(a[i] < ma)ans = min(ans,a[i]*ma);
ma = max(ma , a[i]);
}
}
if(ans == 4e18)puts("IMPOSSIBLE");
else printf("%lld\n",ans);
}
return 0;
}
k Right Expansion Of The Mind
Meaning of the title: a string consists of sss string andttt string, t string is connected to the s string infinitely. If A string is a substring of B string and B string is also a substring of A string, then it can be divided into one group. Output how many groups can be divided, and output the grouping scheme.
Idea: If the letter types of the t string are different, then they must not be grouped together. The s string is traversed backwards to find the first position. The letter at this position is not in the s string. Cut it apart. If the previous part is cut to be exactly the same, then it can be divided into a group.
#include <bits/stdc++.h>
#define ll long long
#define pi pair<int,int>
#define mk make_pair
#define pb push_back
using namespace std;
const int maxn = 1e5+10;
const int mod = 1e9+7,mod1 = 998244353;
const int base = 73,base1 = 47;
struct node {
int sta,val,val1,id;
bool operator < (const node & t)const
{
if(val == t.val){
if(val1 == t.val1)return sta < t.sta;
else return val1 < t.val1;
}
return val < t.val;
}
}a[maxn];
vector<int>G[maxn];
int main()
{
int n;
cin >> n;
for(int i=1;i<=n;i++)
{
string s,s1;
cin >> s >> s1;
int stas = 0;
for(int j=0;j<s1.size();j++)
{
int p = s1[j] - 'a';
stas |= (1<<p);
}
a[i].sta = stas;
a[i].id = i;
int flag = -1;
for(int j=s.size()-1;j>=0;j--)
{
int p = s[j] - 'a';
if((stas&(1<<p)) == 0)
{
flag = j;
break;
}
}
int v = 0,v1 = 0;
for(int j=0;j<=flag;j++)
{
int p = s[j]-'a'+1;
v = 1ll*base*v%mod + p;
v%=mod;
v1 = 1ll*base1*v1%mod1 + p;
v1%=mod1;
}
a[i].val = v;
a[i].val1 = v1;
}
sort(a+1,a+1+n);
int l = 1,r = 1,cnt = 1;
while(r <= n)
{
if(a[l].val == a[r].val && a[l].val1 == a[r].val1 && a[l].sta == a[r].sta)
{
G[cnt].pb(a[r].id);
r++;
}
else l = r,cnt++;
}
cout<<cnt<<'\n';
for(int i=1;i<=cnt;i++)
{
cout<<G[i].size()<<' ';
for(int j=0;j<G[i].size();j++)printf("%d%c",G[i][j],j==G[i].size()-1?'\n':' ');
}
return 0;
}