Problem A?All Palindrome

https://ac.nowcoder.com/acm/contest/624/A

題解：這個字符串只能是所有字符相同的

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; char str[N]; struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);//scanf("%d",&t);//while(t--){scanf("%d",&n);cin>>str;for(int i=0;i<n;i++){a[str[i]-'a']++;}for(int i=0;i<26;i++){ans=max(ans,a[i]);}cout<<n-ans<<endl;//}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

Problem B?Clarifications

https://ac.nowcoder.com/acm/contest/624/B

題解：

C++版本一

#include<bits/stdc++.h> using namespace std; int n,p; int vis[10000]={0}; string pp[110],ss[10000]; int main(){cin>>n>>p;getchar();for(int i=0;i<p;i++){getline(cin,pp[i]);}for(int i=0;i<n;i++){getline(cin,ss[i]);if(ss[i][ss[i].size()-1]=='.'){printf("No Response.\n");continue;}int flag=0;for(int j=0;j<p;j++){if(ss[i]==pp[j]){printf("42.\n");flag=1;break;}}if(flag)continue;vis[i]=1;for(int j=i-1;j>=0;j--){if(ss[j]==ss[i]){vis[i]=vis[j]+1;break;}}if(vis[i]<=5)printf("Read the problem statement carefully.\n");else printf("Juries are investigating. Thanks.\n");}return 0; }

Problem C?∞

https://ac.nowcoder.com/acm/contest/624/C

題解：模擬

1、暴力查找n所在的數是幾位數

2、n所在的數是1中確定的幾位數的第幾個

3、n是2確定的數的第幾位

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; ll t,n,m,k,p,l,r,u,v; ll ans,cnt,flag,temp,sum; int a[N]; char str; struct node{}; ll power(ll a ,ll b){ll res=1;ll base=a;while(b){if(b&1)res=res*base;base=base*base;b>>=1;}return res; } int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);scanf("%lld",&t);while(t--){scanf("%lld",&n);ll l=0;ll r=9;cnt=1;while(1){if(l<n&&n<=r)break;l=r;cnt++;r=r+9*power(10,cnt-1)*cnt;}//cout<<l<<" "<<r<<endl;ll L=1;ll R=9*power(10,cnt-1);ans=1;while(L<=R){ll mid=(L+R)>>1;//cout<<L<<" "<<R<<endl;if(l+(mid-1)*cnt<n&&n<=l+(mid)*cnt){ans=mid;break;}else if(l+(mid-1)*cnt>=n){R=mid-1;}else{L=mid+1;}}//cout<<ans<<endl;ll num=power(10,cnt-1)+ans-1;l=l+(ans-1)*cnt;//cout<<num<<endl;ans=n-l;ans=num/(ll)power(10,cnt-ans)%10;cout<<ans<<endl;}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

Problem G?Virtual Youtuber

https://ac.nowcoder.com/acm/contest/624/G

題解：差分法

求相反

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; ll t,n,m,k,p,l,r,u,v; ll ans,cnt,flag,temp,sum; ll a[N]; char str; struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);scanf("%lld",&t);int T=0;while(t--){scanf("%lld%lld",&n,&p);for(int i=1;i<=n;i++){scanf("%lld",&a[i]);}l=1;r=1;ans=0;temp=a[1];while(r<=n){//cout<<l<<" "<<r<<" "<<temp<<endl;if(temp>=p){ans+=n-r+1;temp-=a[l++];}else{temp+=a[++r];}}cout<<"Case #"<<++T<<": "<<n*(n+1)/2-ans<<endl;}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

Problem J?Bubble Sort

https://ac.nowcoder.com/acm/contest/624/J

題解：乘法逆元 規律

1、交換次數的權呈現正態分布；

2、最少交換次數0，最多交換次數n(n-1)/2；

3、根據數列求和可知答案為n(n-1)/4；

4、費馬小定理

5、快速冪

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=998244353; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; ll t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; char str; struct node{}; ll power(ll a ,ll b,ll c){ll res=1;ll base=a%c;while(b){if(b&1)res=(res*base)%c;base=(base*base)%c;b>>=1;}return res; } int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);scanf("%lld",&t);while(t--){scanf("%lld",&n);printf("%lld\n",((((n%MOD)*((n-1)%MOD))%MOD)*power(4,MOD-2,MOD))%MOD);}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

?

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