Problem A All Palindrome
https://ac.nowcoder.com/acm/contest/624/A
题解：这个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 DEBUG
freopen("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 DEBUG
printf("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
题解：
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
题解：模拟
1、暴力查找n所在的数是几位数
2、n所在的数是1中确定的几位数的第几个
3、n是2确定的数的第几位
/*
*@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 DEBUG
freopen("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 DEBUG
printf("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
题解：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 DEBUG
freopen("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 DEBUG
printf("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
题解：乘法逆元 规律
1、交换次数的权呈现正态分布；
2、最少交换次数0，最多交换次数n(n-1)/2；
3、根据数列求和可知答案为n(n-1)/4；
4、费马小定理
5、快速幂
/*
*@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 DEBUG
freopen("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 DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
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