Codeforces Round #606 (Div. 2, based on Technocup 2020 Elimination Round 4)

由于备战考试这里先挖坑

A. Happy Birthday, Polycarp! (CF 1277 A)

题目大意

解题思路

（位数-1）*9+首位-1+（n是否大于那个值）

神奇的代码

#include <bits/stdc++.h>
#define MIN(a,b) ((((a)<(b)?(a):(b))))
#define MAX(a,b) ((((a)>(b)?(a):(b))))
#define ABS(a) ((((a)>0?(a):-(a))))
using namespace std;

template <typename T>
void read(T &x) {
    int s = 0, c = getchar();
    x = 0;
    while (isspace(c)) c = getchar();
    if (c == 45) s = 1, c = getchar();
    while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
    if (s) x = -x;
}

template <typename T>
void write(T x, char c = ' ') {
    int b[40], l = 0;
    if (x < 0) putchar(45), x = -x;
    while (x > 0) b[l++] = x % 10, x /= 10;
    if (!l) putchar(48);
    while (l) putchar(b[--l] | 48);
    putchar(c);
}

void Input(void) {
    int ans=0;
    int n,m;
    int qwq[22];
    read(n);
    m=n;
    int cnt=0;
    while(n>9){
        m=MIN(m,n%10);
        ans+=9;
        qwq[++cnt]=n%10;
        n/=10;
    }
    ans+=n-1;
    qwq[0]=10;
    for(int i=cnt;i>=0;--i) if (qwq[i]>n){++ans;break;} else if (qwq[i]<n) break;
    write(ans,'\n');
}

void Solve(void) {}

void Output(void) {}

main(void) {
    int kase;
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    read(kase);
    for (int i = 1; i <= kase; i++) {
        //printf("Case #%d: ", i);
        Input();
        Solve();
        Output();
    }
}

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B. Make Them Odd (CF 1277 B)

题目大意

解题思路

记录偶数，去重，把每一个偶数一直除以2得到奇数，记录得到该奇数的最大操作次数，最后是出现的所有奇数的操作次数和。

神奇的代码

#include <bits/stdc++.h>
#define MIN(a,b) ((((a)<(b)?(a):(b))))
#define MAX(a,b) ((((a)>(b)?(a):(b))))
#define ABS(a) ((((a)>0?(a):-(a))))
using namespace std;

template <typename T>
void read(T &x) {
    int s = 0, c = getchar();
    x = 0;
    while (isspace(c)) c = getchar();
    if (c == 45) s = 1, c = getchar();
    while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
    if (s) x = -x;
}

template <typename T>
void write(T x, char c = ' ') {
    int b[40], l = 0;
    if (x < 0) putchar(45), x = -x;
    while (x > 0) b[l++] = x % 10, x /= 10;
    if (!l) putchar(48);
    while (l) putchar(b[--l] | 48);
    putchar(c);
}

unordered_map<int,int> sign;

const int N=2e5+8;

int n,num[N],cnt;

long long ans,an[N];

void Input(void) {
    sign.clear();
    cnt=0;
    read(n);
    for(int u,i=1;i<=n;++i){
        read(u);
        if (((u&1)==0)&&sign[u]==0) {sign[u]=1; num[++cnt]=u;} 
    }
}

void Solve(void) {
    ans=0;
    sort(num+1,num+1+cnt);
    for(int u,cout,i=cnt;i>=1;--i)
        if (sign[num[i]]){
            u=num[i];
            u>>=1;
            cout=1;
            while((u&1)==0){
                if (sign[u]) sign[u]=0;
                u>>=1;
                ++cout;
            }
            an[i]=cout;
        }
    for(int i=cnt;i>=1;--i)
        if (sign[num[i]]) ans+=an[i];
}

void Output(void) {
    write(ans,'\n');
}

main(void) {
    int kase;
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    read(kase);
    for (int i = 1; i <= kase; i++) {
        //printf("Case #%d: ", i);
        Input();
        Solve();
        Output();
    }
}

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C. As Simple as One and Two (CF 1277 C)

题目大意

解题思路

先检查是否有twone，把o去掉，然后再检查one和two，把中间字母去掉即可。

神奇的代码

#include <bits/stdc++.h>
#define MIN(a,b) ((((a)<(b)?(a):(b))))
#define MAX(a,b) ((((a)>(b)?(a):(b))))
#define ABS(a) ((((a)>0?(a):-(a))))
using namespace std;

template <typename T>
void read(T &x) {
    int s = 0, c = getchar();
    x = 0;
    while (isspace(c)) c = getchar();
    if (c == 45) s = 1, c = getchar();
    while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
    if (s) x = -x;
}

template <typename T>
void write(T x, char c = ' ') {
    int b[40], l = 0;
    if (x < 0) putchar(45), x = -x;
    while (x > 0) b[l++] = x % 10, x /= 10;
    if (!l) putchar(48);
    while (l) putchar(b[--l] | 48);
    putchar(c);
}

char one[4]={"one"};

char two[4]={"two"};

char twone[6]={"twone"};

int nxtone[4],nxttwo[4],nxttwone[6];

const int N=2e5+8;

char s[N];

int len,cnt;

int anspos[N];

void makenext(char qaq[],int nxt[],int n){
    nxt[0]=0;
    int k=0;
    for (int i=1;i<n;i++){
        while ((k>0)&&(qaq[k]!=qaq[i])) k=nxt[k-1];
        if (qaq[k]==qaq[i]) k++;
        nxt[i]=k;
    }
}

void kmp(char qaq[],int nxt[],char qwq[],int n){
    int q=0;
    for (int i=0;i<len;i++){
        if (qwq[i]=='@') continue;
        while ((qwq[i]!=qaq[q])&&(q>0)) q=nxt[q-1];   
        if (qwq[i]==qaq[q]) q++;
        if (q==n) {
            if (n==5) {qwq[i-2]='@'; anspos[++cnt]=i-1;}
            else {qwq[i-1]='@'; anspos[++cnt]=i;}
        }
    }
}

void Input(void) {
    cnt=0;
    scanf("%s",s);
    len=strlen(s);
}

void Solve(void) {
    kmp(twone,nxttwone,s,5);
    kmp(two,nxttwo,s,3);
    kmp(one,nxtone,s,3);
}

void Output(void) {
    write(cnt,'\n');
    for(int i=1;i<=cnt;++i) printf("%d%c",anspos[i],i==cnt?'\n':' ');
    if (cnt==0) puts("");
}

main(void) {
    int kase;
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    read(kase);
    makenext(one,nxtone,3);
    makenext(two,nxttwo,3);
    makenext(twone,nxttwone,5);
    for (int i = 1; i <= kase; i++) {
        //printf("Case #%d: ", i);
        Input();
        Solve();
        Output();
    }
}

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D. Let's Play the Words? (CF 1277 D)

题目大意

解题思路

每个01串按照首位和末位进行分类，共四类，-1的情况即是仅有00和11而无01或10，注意到如果有01或10串翻转得到另一个已有的串，则这两个本身可以搭配，可以忽略，然后统计01和10的数量，翻转的最小的次数即是两者数量差值除以2.

神奇的代码

#include <bits/stdc++.h>
#define MIN(a,b) ((((a)<(b)?(a):(b))))
#define MAX(a,b) ((((a)>(b)?(a):(b))))
#define ABS(a) ((((a)>0?(a):-(a))))
using namespace std;

template <typename T>
void read(T &x) {
    int s = 0, c = getchar();
    x = 0;
    while (isspace(c)) c = getchar();
    if (c == 45) s = 1, c = getchar();
    while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
    if (s) x = -x;
}

template <typename T>
void write(T x, char c = ' ') {
    int b[40], l = 0;
    if (x < 0) putchar(45), x = -x;
    while (x > 0) b[l++] = x % 10, x /= 10;
    if (!l) putchar(48);
    while (l) putchar(b[--l] | 48);
    putchar(c);
}

const int N=2e5+8;

int num[4],s01[N],s10[N];

int n,l,c0,c1;

string s,ss;

unordered_map<string,pair<int,bool>> qwq;

void Input(void) {
    qwq.clear();
    num[0]=num[1]=num[2]=num[3]=0;
    c0=c1=0;
    cin>>n;
    for(int i=1;i<=n;++i){
        cin>>s;
        if (s[0]==s[s.size()-1]) {qwq[s]=make_pair(i,1);continue;}
        ss=s;
        reverse(s.begin(),s.end());
        if (qwq[s].second==1) {qwq[s].second=0;++num[2];++num[3];continue;}
        qwq[ss]=make_pair(i,1);
    }
}

void Solve(void) {
    for(auto v:qwq){
        if (v.second.second==1){
        const char *ss=v.first.c_str();
            l=strlen(ss);
            if (l==1) ++num[ss[0]-'0'];
            else{
                if (ss[0]=='0'&&ss[l-1]=='0') ++num[0];
                else if (ss[0]=='0'&&ss[l-1]=='1') ++num[2],s01[++c0]=v.second.first;
                else if (ss[0]=='1'&&ss[l-1]=='0') ++num[3],s10[++c1]=v.second.first;
                else ++num[1];
            }
        }
    }
    if (num[2]==0&&num[3]==0&&num[0]!=0&&num[1]!=0) cout<<"-1"<<endl;
    else {
        int qwq=ABS(num[2]-num[3]);
        qwq/=2;
        cout<<qwq<<endl;
        if (num[2]>num[3]) for(int i=1;i<=qwq;++i) cout<<s01[i]<<(i==qwq?'\n':' ');
        else for(int i=1;i<=qwq;++i) cout<<s10[i]<<(i==qwq?'\n':' ');
        if (qwq==0) cout<<'\n';
    }
}

void Output(void) {}

main(void) {
    ios::sync_with_stdio(false);
    int kase;
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    cin>>kase;
    for (int i = 1; i <= kase; i++) {
        //printf("Case #%d: ", i);
        Input();
        Solve();
        Output();
    }
}

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