2018ICPC 北京

A.0:27:15 solved by hl

感人的数据范围，每次连上边之后dfs一下前者看能不能找到自己就行

#include <iostream>
#include <string>
#include <map>
#include <vector>
#include <cstdio>
using namespace std;
int N;
const int maxn = 110;
map<string,int>Hash;
vector<int>P[maxn];
bool vis[maxn];
int s;
bool dfs(int t){
    vis[t] = 1;
    for(int i = 0 ; i < P[t].size(); i ++){
        int v = P[t][i];
        if(v == s) return false;
        if(vis[v]) continue;
        if(!dfs(v)) return false;
    }
    return true;
}
int main()
{
   while(~scanf("%d",&N)){
        int cnt; cnt = 0; Hash.clear();
        for(int i = 0 ; i < maxn; i ++) P[i].clear();
        int flag = 0;
        for(int i = 1; i <= N; i ++){
            string a,b; cin >> a >> b;
            if(flag) continue;
            if(!Hash[a]) Hash[a] = ++cnt;
            if(!Hash[b]) Hash[b] = ++cnt;
            P[Hash[a]].push_back(Hash[b]);
            for(int j = 0; j < maxn; j ++) vis[j] = 0;
            s = Hash[a];
            if(!dfs(Hash[a])){
                cout << a << " " << b << endl;
                flag = 1;
            }
        }
        if(!flag) puts("0");
   }
    return 0;
}

B.2:34:44(-6) solved by hl,gbs

debug题,两个人各写了一遍各WA了几发，最后提出0没有前导0的问题让两份代码都过了

先将所有串拼成一个很长的字符串，如果前一行末尾和后一行开头有一个不是数字，就要在两者之间加上空格

拼完之后就是一个简单的按题目统计了，对于每行出现的次数，要用一个数组记录一下每个字符串拼起来之后的断点

#include <iostream>
#include <string>
#include <map>
#include <vector>
#include <cstdio>
#include <sstream>
#include <algorithm>
using namespace std;
int N;
const int maxn = 10010;
string tmp,a;
int id[maxn];
int num[maxn];
int main()
{
    getline(cin,a); int cnt = 1;
    for(int i = 0 ; i < a.size(); i ++){
        if(a[i] != ' ') id[1]++;
    }
    while(getline(cin,tmp)){
        cnt++;
        if(!isdigit(a[a.size() - 1]) || !isdigit(tmp[0])){
            a.push_back(' ');
        }
        id[cnt] = id[cnt - 1];
        for(int i = 0 ; i < tmp.size(); i ++){
            if(tmp[i] != ' ') id[cnt]++;
        }
        a = a + tmp;
    }
   // cout << a << endl;
    stringstream ss(a);
    int sum = 0;
    int fi = 0 ;
    while(ss >> tmp){
        //cout << tmp << endl;
        int flag = 1;
        if(tmp[0] == '0' && tmp.size() > 1) flag = 0;
        if(islower(tmp[0]) || islower(tmp[tmp.size() - 1])) flag = 0;
        if(flag){
            int p = upper_bound(id + 1,id + 1 + cnt,sum) - id;
            num[p]++;
            if(!fi) fi = 1;
            else cout << " ";
            cout << tmp;
        }
        sum += tmp.size();
    }
    cout << endl;
    for(int i = 1; i <= cnt; i ++){
        printf("%d
",num[i]);
    }
    return 0;
}

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
char sum1[1000005];
int ans[3005];

int now_len1 = 0;
int yuan;
char lls1[1000005];
int now_row;
int llen;
bool if_first = true;
inline bool if_d(char a)
{
    if (a>'9')
        return false;
    if (a<'0')
        return false;
    return true;
}
bool if_OK;
void end1()
{

    lls1[llen] = 0;
    //cout<<lls1<<endl;

    if (llen != 0)
    {
        if (lls1[0] == '0' && llen != 1)
        {
            llen= 0;
            return ;
        }
        if (if_d(lls1[0])&& if_d(lls1[llen-1]) )
        {//cout<<"   &&   "<<llen<<endl;
            if (!if_first)
            {
                printf(" ");
            }

            if_first = false;
            printf("%s",lls1);
            ans[yuan]++;
        }
    }
    llen =0;



}

int main()
{
    char a1;
    while(scanf("%c",&a1)!= EOF)
    {

        sum1[now_len1++] = a1;
    }
    //cout<<sum1;
    now_row = 0;
    llen =0;
    memset(ans,0,sizeof(ans));
    for (int i=0;i<now_len1; i++)
    {
        //cout<<now_len1
        if (sum1[i] == ' ')
            end1();
        else if (sum1[i] == '
')
        {
            if (sum1[i+1] != 0)
                now_row++;
            if (i == 0)
                end1();
            else if(if_d(sum1[i-1]) && if_d(sum1[i+1]))
            {
                //cout<<"P"<<endl;
            }
            else
                end1();
        }
        else
        {
            lls1[llen++] = sum1[i];
            if (llen == 1)
            {
                yuan = now_row;
                if_OK = true;
            }
            if (!if_d(sum1[i]))
            {
                if_OK = false;
            }

        }
    }
    end1();
    cout<<endl;
    for (int i=0; i<=now_row; i++)
    {
        cout<<ans[i]<<endl;
    }

    return 0;
}
/*
a19 01 17b
12 bdc 13
23 14 0 bc




*/

D.1:30:06(-1) solved by zcz

是个构造,具体方法看代码

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
using namespace std;
int vis[40];

vector<int> ans[2];

int main()
{
    long long M;
    while(cin>>M)
    {
        if(M==0)
        {
            cout<<2<<endl;
            cout<<"1 1"<<endl;
            cout<<"2 2"<<endl;
            continue;
        }
        ans[0].clear();
        ans[1].clear();
        memset(vis,0,sizeof(vis));
        int l=1;
        int cnt=0;
        while(M)
        {
            if(M&1)
            {
                cnt++;
                vis[l]=1;
            }
            l++;
            M>>=1;
        }
        int t=1;
        int i=0;
        int num=0;
        while(cnt)
        {
            if(vis[t])
            {
                ans[0].push_back(i+1);
                ans[1].push_back(199);
                num++;
                //cout<<i+1<<' '<<199<<endl;
                vis[t]=0;
                cnt--;
                i=i+2;
            }
            else
            {
                ans[0].push_back(i+1);
                ans[1].push_back(i+2);
                num++;
                //cout<<i+1<<' '<<i+2<<endl;
                i=i+2;
                t++;
            }
        }
        ans[0].push_back(i);
        ans[1].push_back(i);
        num++;
        //cout<<i<<' '<<i<<endl;
        cout<<num<<endl;
        for(int j=0;j<num;j++)
        {
            cout<<ans[0][j]<<' '<<ans[1][j]<<endl;
        }
    }


    return 0;
}

H.3:57:53 solved by hl

将所给模式串分成N + 1个模式串

其中一个是原来的，另一个是变了一个字母的

dp[i][j]表示i号模式串在以j位置结尾处出现的次数，每次容斥减掉前面出现过的次数

最后将每个dp[i][j] * 2 ^ (M - j)次方计入贡献，表示前面匹配成功之后后面就可以随便写

9.5补：hl托更了两周的AC自动机标准解法

#include <iostream>
#include <string>
#include <map>
#include <cstring>
#include <vector>
#include <cstdio>
#include <sstream>
#include <algorithm>
#define LL long long
using namespace std;
int N,M;
const int maxn = 110;
char str[maxn][maxn];
LL dp[maxn][maxn];
LL pow2[maxn],sum[maxn];
int main()
{
    int T; scanf("%d",&T);
    for(int i = 0; i <= 40; i ++) pow2[i] = (1LL << i);
    while(T--){
        memset(dp,0,sizeof(dp));
        memset(sum,0,sizeof(sum));
        scanf("%d%d",&N,&M);
        scanf("%s",str[0] + 1);
        dp[0][N] = 1;
        for(int i = 1; i <= N; i ++){
            for(int j = 1; j <= N ; j ++) str[i][j] = str[0][j];
            if(str[i][i] == '0') str[i][i] = '1';
            else str[i][i] = '0';
            dp[i][N] = 1;
        }
         if(M < N){
            puts("0");
            continue;
        }
        sum[N] = (N + 1);
        for(int i = N + 1; i <= M ; i ++){
            for(int j = 0 ; j <= N ; j ++){
                dp[j][i] = pow2[i - N] - sum[i - N];
               // cout << dp[j][i] << endl;
                for(int k = 0; k <= N ; k ++){
                  //  if(k == j) continue;
                    for(int l = 2; l <= N ; l ++){
                        int num = 0;
                        for(int p = l; p <= N && num < 1; p ++){
                            if(str[k][p] != str[j][p - l + 1]) num++;
                        }
                        if(num < 1){
                            dp[j][i] -= dp[k][i - l + 1];
                       //     cout << j << ' ' << k << " " << dp[k][i - l + 1] << " " << i - l + 1 << endl;
                        }

                    }
                }
               // cout << j << " " << i << " " << dp[j][i] << endl;
                sum[i] += dp[j][i];
            }
            sum[i] += sum[i - 1] * 2;
          //  cout << sum[i] << endl;
        }
        LL ans = 0;
        for(int i = 0; i <= N ; i ++){
            for(int j = N ; j <= M ; j ++){
                ans += dp[i][j] * pow2[M - j];
            }
        }
        printf("%lld
",ans);
    }
    return 0;
}

I.1:19:19 solved by gbs

是个找规律，也可以推式子

具体规律看代码

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
/*
int anz[45];
int an0[45];



void dfs(int n)
{
    int lln1 =n;
    INT T1;

    for (int i=0; i<20; i++)
    {
        if (lln1 <=anz[i])
        {
            if (i == 0)
                cout<<lln1-1;
            else if (i == 1)
                cout<<lln1<<lln1;
            else
            {
                t1 =lln1/anz[i-2];
                t1++;
                cout<<t1;
                ans[now_b++] = t1;
                dfs();

            }
            return;
        }
        lln1 -= anz[i];
    }

}*/
char str1[100005];
int ans[200005];
int now_b;
int main()
{
    /*anz[0] = an0[0] = 10;
    anz[1] = 9;
    an0[1] = 10;
    for (int i=2; i<20; i++)
    {
        anz[i] = an0[i-2]*9;
        an0[i] = an0[i-2]*10;
        cout<<anz[i]<<endl;
    }*/
    int T;
    //int n;
    cin >>T;
    while(T--)
    {
        scanf("%s",str1);
        int len1 = strlen(str1);
        int t1;
        if (len1 == 1)
        {
            t1 = str1[0]-'0';
            t1--;
            cout<<t1<<endl;
            continue;
        }
        if (len1 == 2 &&str1[0] == '1'&&str1[1]=='0' )
        {
            cout<<9<<endl;
            continue;
        }
        if (str1[0] == '1'&&str1[1]>='1' )
        {
            t1 = str1[1] -'0';
            printf("%d",t1);
            for (int i=2; i<len1; i++)
                printf("%c",str1[i]);
            for (int i=len1-1; i>=2; i--)
                printf("%c",str1[i]);
            printf("%d
",t1);
            continue;
        }
        if (str1[0] == '1' )
        {
            t1 = 9;
            printf("%d",t1);
            for (int i=2; i<len1; i++)
                printf("%c",str1[i]);
            for (int i=len1-2; i>=2; i--)
                printf("%c",str1[i]);
            printf("%d
",t1);
            continue;
        }
        else
        {
            t1 = str1[0]-'1';
            printf("%d",t1);
            for (int i=1; i<len1; i++)
                printf("%c",str1[i]);
            for (int i=len1-2; i>=1; i--)
                printf("%c",str1[i]);
            printf("%d
",t1);
            continue;
        }


    }
    return 0;
}